DP... dp[k][i][j]: number of subsequences using first k elements, where the last two parities are (i, j), with

include <bits/stdc++.h>

using namespace std;

static const long long MOD = 1000000007;

int main(){ ios::sync_with_stdio(false); cin.tie(nullptr);

int n;
cin >> n;
vector<long long> a(n);
for(int i = 0; i < n; i++){
    cin >> a[i];
}

// dp[k][i][j]: number of subsequences using first k elements
// where the last two parities are (i, j), with:
// 0 = none, 1 = even, 2 = odd
vector<vector<vector<long long>>> dp(n + 1, vector<vector<long long>>(3, vector<long long>(3, 0)));
dp[0][0][0] = 1; // Base case: empty subsequence

for(int k = 0; k < n; k++){
    int p = (a[k] % 2 == 0 ? 1 : 2);
    for(int i = 0; i < 3; i++){
        for(int j = 0; j < 3; j++){
            // Case 1: don't take a[k]
            dp[k + 1][i][j] = (dp[k + 1][i][j] + dp[k][i][j]) % MOD;

            // Case 2: take a[k] if it doesn't form 3 same parity
            if (!(i == j && j == p)) {
                dp[k + 1][j][p] = (dp[k + 1][j][p] + dp[k][i][j]) % MOD;
            }
        }
    }
}

// sum over all final dp[n][i][j] and subtract empty subsequence
long long ans = 0;
for(int i = 0; i < 3; i++){
    for(int j = 0; j < 3; j++){
        ans = (ans + dp[n][i][j]) % MOD;
    }
}
ans = (ans - 1 + MOD) % MOD;

cout << ans << "\n";
return 0;

}