The presidential election is coming in Bearland next year! Everybody is so excited about this!

So far, there are three candidates, Alice, Bob, and Charlie. 

There are n citizens in Bearland. The election result will determine the life of all citizens of Bearland for many years. Because of this great responsibility, each of n citizens will choose one of six orders of preference between Alice, Bob and Charlie uniformly at random, independently from other voters.

The government of Bearland has devised a function to help determine the outcome of the election given the voters preferences. More specifically, the function is <image> (takes n boolean numbers and returns a boolean number). The function also obeys the following property: f(1 - x1, 1 - x2, ..., 1 - xn) = 1 - f(x1, x2, ..., xn).

Three rounds will be run between each pair of candidates: Alice and Bob, Bob and Charlie, Charlie and Alice. In each round, xi will be equal to 1, if i-th citizen prefers the first candidate to second in this round, and 0 otherwise. After this, y = f(x1, x2, ..., xn) will be calculated. If y = 1, the first candidate will be declared as winner in this round. If y = 0, the second will be the winner, respectively.

Define the probability that there is a candidate who won two rounds as p. p·6n is always an integer. Print the value of this integer modulo 109 + 7 = 1 000 000 007.

Input

The first line contains one integer n (1 ≤ n ≤ 20).

The next line contains a string of length 2n of zeros and ones, representing function f. Let bk(x) the k-th bit in binary representation of x, i-th (0-based) digit of this string shows the return value of f(b1(i), b2(i), ..., bn(i)).

It is guaranteed that f(1 - x1, 1 - x2, ..., 1 - xn) = 1 - f(x1, x2, ..., xn) for any values of x1, x2, ldots, xn.

Output

Output one integer — answer to the problem.

Examples

Input

3
01010101


Output

216


Input

3
01101001


Output

168

Note

In first sample, result is always fully determined by the first voter. In other words, f(x1, x2, x3) = x1. Thus, any no matter what happens, there will be a candidate who won two rounds (more specifically, the candidate who is at the top of voter 1's preference list), so p = 1, and we print 1·63 = 216.
#include <bits/stdc++.h>
using namespace std;
const int N = 20;
const int Q = 1e9 + 7;
int n, f[1 << N], cnt[1 << N];
char str[(1 << N) + 5];
int power[7][N + 1];
int inverse(int x) {
  return x == 1 ? x : (long long)(Q - Q / x) * inverse(Q % x) % Q;
}
void FWT(int a[], int len, int oper) {
  for (int k = 0; 1 << k < len; ++k) {
    for (int i = 0; i < len; ++i) {
      if (~i >> k & 1) {
        int j = i ^ (1 << k);
        int x = (a[i] + Q - a[j]) % Q;
        int y = (a[i] + a[j]) % Q;
        if (oper == -1) {
          x = (Q - x) % Q;
          swap(x, y);
        }
        a[i] = x;
        a[j] = y;
      }
    }
  }
}
int main() {
  scanf("%d%s", &n, str);
  for (int i = 0; i <= 6; ++i) {
    power[i][0] = 1;
    for (int j = 1; j <= n; ++j) {
      power[i][j] = (long long)power[i][j - 1] * i % Q;
    }
  }
  for (int i = 0; i < 1 << n; ++i) {
    f[i] = str[i] - '0';
  }
  for (int i = 1; i < 1 << n; ++i) {
    cnt[i] = cnt[i & (i - 1)] + 1;
  }
  FWT(f, 1 << n, 1);
  for (int i = 0; i < 1 << n; ++i) {
    f[i] = (long long)f[i] * f[i] % Q;
  }
  FWT(f, 1 << n, -1);
  int inv = inverse(1 << n);
  for (int i = 0; i < 1 << n; ++i) {
    f[i] = (long long)f[i] * inv % Q;
  }
  int res = 0;
  for (int i = 0; i < 1 << n; ++i) {
    res += (long long)f[i] * power[1][cnt[i]] % Q * power[2][n - cnt[i]] % Q;
    res %= Q;
  }
  res = res * 3LL % Q;
  cout << res << endl;
}
