Read the title of the Blog. There exists symmetry in our problem since every final string is equal likely to be the final string.

Assume every $$$?$$$ is replaced by $$$a$$$, $$$b$$$, or $$$c$$$ uniformly at random, each with probability $$$\frac{1}{3}$$$. By Linearity of Expectation, we can iterate over every index $$$i$$$ and ask: What is the expected number of $$$abc$$$ subsequences where $$$s[i]$$$ acts as the middle character $$$b$$$?

Let us count the Expected value of having $$$abc$$$ as a subsequence and then multiply by $$$3^{\text{count of '?'}}$$$. For a fixed middle index $$$i$$$, the expected number of valid 'abc's passing through it is the product of three independent probabilities/expectations

The Left Side ($$$E[A_{left}]$$$): Every fixed 'a' to the left contributes exactly 1. Every '?' to the left has a $$$\frac{1}{3}$$$ chance of becoming an 'a'.

The Right Side ($$$E[C_{right}]$$$): Similarly, every fixed 'c' to the right contributes 1, and every '?' contributes $$$\frac{1}{3}$$$.

The Middle ($$$P(s[i] == \text{'b'})$$$): If $$$s[i]$$$ is 'b', this probability is 1. If $$$s[i]$$$ is '?', the probability is $$$\frac{1}{3}$$$. If it's 'a' or 'c', it's 0.

Because the choices for the '?' characters on the left, the right, and the middle are strictly independent, the expected value of their product is the product of their expected values:

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast, unroll-loops")
#ifdef ONLINE_JUDGE
    #define debug(...) ((void)0)
#else
    #include "debug.h"
#endif
using namespace std;
using ll = long long;
#define endl '\n'

template <int MOD_> struct modnum {
	static constexpr int MOD = MOD_;
	static_assert(MOD_ > 0, "MOD must be positive");
private:
	static int minv(int a, int m) {
		a %= m;
		assert(a);
		return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
	}
public:
	int v;
	modnum() : v(0) {}
	modnum(ll v_) : v(int(v_ % MOD)) { if (v < 0) v += MOD; }
	explicit operator int() const { return v; }
	friend std::ostream& operator << (std::ostream& out, const modnum& n) { return out << int(n); }
	friend std::istream& operator >> (std::istream& in, modnum& n) { ll v_; in >> v_; n = modnum(v_); return in; }
	friend bool operator == (const modnum& a, const modnum& b) { return a.v == b.v; }
	friend bool operator != (const modnum& a, const modnum& b) { return a.v != b.v; }
	modnum inv() const { modnum res; res.v = minv(v, MOD); return res; }
	friend modnum inv(const modnum& m) { return m.inv(); }
	modnum neg() const { modnum res; res.v = v ? MOD-v : 0; return res; }
	friend modnum neg(const modnum& m) { return m.neg(); }
	modnum operator- () const { return neg(); }
	modnum operator+ () const { return modnum(*this); }
	modnum& operator ++ () { v ++; if (v == MOD) v = 0; return *this; }
	modnum& operator -- () { if (v == 0) v = MOD; v --; return *this; }
	modnum& operator += (const modnum& o) { v += o.v; if (v >= MOD) v -= MOD; return *this; }
	modnum& operator -= (const modnum& o) { v -= o.v; if (v < 0) v += MOD; return *this; }
	modnum& operator *= (const modnum& o) { v = int(ll(v) * ll(o.v) % MOD); return *this; }
	modnum& operator /= (const modnum& o) { return *this *= o.inv(); }
	friend modnum operator ++ (modnum& a, int) { modnum r = a; ++a; return r; }
	friend modnum operator -- (modnum& a, int) { modnum r = a; --a; return r; }
	friend modnum operator + (const modnum& a, const modnum& b) { return modnum(a) += b; }
	friend modnum operator - (const modnum& a, const modnum& b) { return modnum(a) -= b; }
	friend modnum operator * (const modnum& a, const modnum& b) { return modnum(a) *= b; }
	friend modnum operator / (const modnum& a, const modnum& b) { return modnum(a) /= b; }
};

const int MOD = 1e9 + 7;
using mint = modnum<MOD>;
template <typename T> mint binpow(T aa, long long b) {
	assert(b >= 0);
	mint a = aa;
	mint r = 1; while (b) { if (b & 1) r *= a; b >>= 1; a *= a; } return r;
}

template <int N> 
struct counting{
	mint fact[N], inv[N], invfact[N];
	counting() {
		fact[0] = fact[1] = invfact[0] = invfact[1] = inv[1] = 1;
		for (int i = 2; i < N; i++) {
			fact[i] = mint(i) * fact[i - 1];
			inv[i] = mint(0) - mint(MOD / i) * inv[MOD % i];
			invfact[i] = invfact[i - 1] * inv[i];
		}
	}
	mint choose(int n, int r) {
		if (n < 0 || r < 0 || n < r) return mint(0);
		return fact[n] * invfact[n - r] * invfact[r];
	}
	mint stars_and_bars(int n, int k){
		return choose(n + k - 1, k - 1);
	}
	mint P(int n, int r){
        if (n < 0 || r < 0 || r > n) return mint(0);
        return fact[n] * invfact[n - r];
    }
	mint catalan(int n){
        return fact[2 * n] * invfact[n] * invfact[n + 1];
    }
};
const int N = 1e6 + 5;
// counting<N> comb;

// DO NOT FORGET TO CHANGE THE MOD
void SOLVE() {
    int n; cin >> n;
    string s; cin >> s;
	mint answer = 0;
    mint A = 0;
    mint C = count(begin(s), end(s), 'c');
    mint markL = 0, markR = count(begin(s), end(s), '?');
    mint INV = mint(3).inv();

    for(int i = 0; i < n; i++){
        if(s[i] == 'c') C--;
        else if(s[i] == '?') markR--;
        if(s[i] == 'b' || s[i] == '?') {
            answer += (A + INV * markL) * (C + INV * markR) * (s[i] == '?' ? INV : mint(1));
        }
        if(s[i] == 'a') A++;
        else if(s[i] == '?') markL++;
    }
    cout << answer * binpow(3, (int)markL);
}
int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr); std::cout.tie(nullptr);
	// int o_o = 1; cin >> o_o; while(o_o --)
	SOLVE(); return 0;
}

If you try to calculate the probabilities directly, you will quickly realize that the seats are not equal. A person aiming for seat $$$1$$$ and walking left falls off the plane immediately if it's occupied, while a person walking right has $$$n-1$$$ backup options. The edges of the array completely destroy any natural symmetry. But what if there were no edges? How can we make every single position perfectly symmetric to all the others?

To remove the edges, we need to turn our linear airplane into a circle. Imagine we add exactly one dummy seat at the very end of the plane, giving us $$$n+1$$$ seats. Then, we connect this dummy seat back to the first seat. Now, passengers walk in a continuous loop. No one ever gets angry or falls off because they will just keep circling until they find an empty seat. In this circular universe, a boarding configuration is valid for our original straight airplane if and only if the dummy seat at $$$n+1$$$ remains completely empty. If someone had to sit in the dummy seat, it means they walked off the edge of the real plane.

Read the two hints first.

There are now $$$n+1$$$ seats. Each of the $$$m$$$ passengers independently chooses one of these $$$n+1$$$ seats and one of $$$2$$$ directions. This gives us a total of $$$(2(n+1))^m$$$ possible starting configurations. Since there are $$$n+1$$$ total seats and $$$m$$$ passengers will eventually sit down, there will always be exactly $$$n+1-m$$$ empty seats left over. I think if you reach this point of the blog I can left this part as an exercise for the reader of calculating the probability of having the seat $$$n+1$$$ empty.

To find the answer to the original problem, we just take the total number of configurations in our circular universe and multiply it by the probability that the configuration is actually valid, $$$(2(n+1))^m \times \frac{n+1-m}{n+1}$$$.