First thing to notice is that there are not a lot of different gcd's at all. Number of possible gcds is log2(10^10).

#include <stdio.h>
 
unsigned long long n, res, argres;
int p, primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 51, 53, 59, 61, 67, 71};


unsigned long long mul(unsigned long long a, unsigned long long b){
    unsigned long long res = 0;
 
    while (b){
        if (b & 1LL) res = (res + a);
        if (res >= n) return 0;
        a = (a << 1LL);
        b >>= 1LL;
    }
 
    return res;
}
 
void backtrack(int i, int lim, unsigned long long val, unsigned long long r){
    if (r > res) {
        argres = val;
        res = r;
    }
    if (i == p) return;
 
    int d;
    unsigned long long x = val;
 
    for (d = 1; d <= lim; d++){
        x = mul(x, primes[i]);
        if (x == 0) return;
        backtrack(i + 1, d, x, r * (d + 1));
    }
}
 
int main(){
    /* Tested for n <= 10^18 */
 
    p = sizeof(primes) / sizeof(int);
 
    while (scanf("%llu", &n) != EOF){
        res = 0;
        backtrack(0, 100, 1, 1);
        printf("Maximum number of divisors of any number less than %llu = %llu, argres==%llu\n", n, res, argres);
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

#define DEBUG 1

void debug(const char* p){
    #if DEBUG
    freopen(p, "r", stdin); 
    #else
    fastio;
    #endif      
} 

signed main(int argc, char** argv){
    debug("test1.txt");
    ll n, k;
    cin >> n >> k;
    map<ll, ll> indices;
    for(ll i = 1, tmp; i <= n; ++i){
        cin >> tmp;
        for(ll j = 1; j * j <= tmp; ++j){
            if(tmp%j == 0){
                indices[j]++;
                if(tmp/j != j){
                    indices[tmp/j]++;
                }
            }
        }
    }
    ll ans = 0;
    for(const auto& [factor, times]: indices){
        //printf("factor==%lld, times==%lld\n", factor, times);
        if(times >= k){
            ans = max(ans, factor);
        }
    }
    cout << ans << "\n";
}