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 maxn = 1048576 + 77;
const int mod = 1E9 + 7;
const int divv = 500000004;
void FWT(int arr[], int len, long long p) {
  int n = 1 << len;
  for (int i = 0; i < len; ++i) {
    int m = 1 << i;
    for (int j = 0; j < n; j += (m << 1)) {
      for (int k = 0; k < m; ++k) {
        int p1 = j + k, p2 = j + m + k, u = arr[p1], t = arr[p2];
        arr[p1] = u + t;
        if (arr[p1] >= mod) arr[p1] -= mod;
        arr[p2] = u - t;
        if (arr[p2] < 0) arr[p2] += mod;
        if (p != 1) {
          arr[p1] = (arr[p1] * p) % mod;
          arr[p2] = (arr[p2] * p) % mod;
        }
      }
    }
  }
}
int N, A[maxn];
char S[maxn];
int main() {
  scanf("%d%s", &N, S);
  int tot = 1 << N;
  for (int i = 0; i < tot; ++i) A[i] = S[i] - '0';
  FWT(A, N, 1);
  for (int i = 0; i < tot; ++i) A[i] = (A[i] * (long long)A[i]) % mod;
  FWT(A, N, divv);
  long long ret = 0;
  for (int i = 0; i < tot; ++i) {
    ret += ((tot >> __builtin_popcount(i)) * (long long)A[i]) % mod;
    if (ret >= mod) ret -= mod;
  }
  ret = (ret * 3) % mod;
  cout << ret << endl;
}
