#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <cmath>
#include <numeric>
#include <functional>
#include <cassert>
#include <atcoder/modint>

#define debug_value(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << #x << "=" << x << endl;
#define debug(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << x << endl;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using namespace std;
typedef long long ll;

template<typename T>
vector<vector<T>> vec2d(int n, int m, T v){
    return vector<vector<T>>(n, vector<T>(m, v));
}

template<typename T>
vector<vector<vector<T>>> vec3d(int n, int m, int k, T v){
    return vector<vector<vector<T>>>(n, vector<vector<T>>(m, vector<T>(k, v)));
}

template<typename T>
void print_vector(vector<T> v, char delimiter=' '){
    if(v.empty()) {
        cout << endl;
        return;
    }
    for(int i = 0; i+1 < v.size(); i++) cout << v[i] << delimiter;
    cout << v.back() << endl;
}

using mint = atcoder::modint1000000007;

ostream& operator<<(ostream& os, const mint& m){
    os << m.val();
    return os;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    ll n; cin >> n;
    string k; cin >> k;
    mint ans = 0;
    if(n == 0){
        for(int i = 0; i < 8; i++){
            if(k[i] == '-') continue;
            ans += 1;
        }
        cout << ans << endl;
        return 0;
    }
  
    for(int i = 0; i < 8; i++){
        if(k[i] == '-') continue;
        int c = __builtin_popcount(i);
        mint ave = mint(c)/4;
        mint ave_other = 1-ave;
        mint m = mint(4).pow(n-1);
        mint sum = mint(2).pow(n-1)-1;
        ans += mint(4).pow(n)*(mint(2).pow(n-1)+1);
        { // p s_i qの最上位が1
            ans += (ave-ave_other)*sum*m*c;
        }
        { // p s_i qの最上位が0
            ans += (ave_other-ave)*sum*m*(4-c);
        }
    }
    cout << ans << endl;
}