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 MOD = 1e9 + 7;
const int INV_2 = 5e8 + 4;
const int SIZE = (1 << 20) + 5;
char s[SIZE];
int a[1 << 21], res[1 << 21];
inline void FWT(int c[], int N, int tf_utf) {
  for (int i = 1; i < N; i <<= 1) {
    int tmp = i << 1;
    for (int j = 0; j < N; j += tmp) {
      for (int k = 0; k < i; k++) {
        int x = c[j + k], y = c[j + k + i];
        if (tf_utf) {
          c[j + k] = x + y;
          if (c[j + k] >= MOD) {
            c[j + k] -= MOD;
          }
          c[j + k + i] = x - y;
          if (c[j + k + i] < 0) {
            c[j + k + i] += MOD;
          }
        } else {
          c[j + k] = (long long)(x + y) * INV_2 % MOD;
          c[j + k + i] = (long long)(x - y + MOD) * INV_2 % MOD;
        }
      }
    }
  }
}
int main() {
  int(n);
  scanf("%d", &n);
  scanf("%s", (s));
  for (int i = 0; i < (1 << n); ++i) {
    a[(i << 1) + (s[i] - '0')] = 1;
  }
  FWT(a, 1 << (n + 1), 1);
  for (int i = 0; i < (1 << (n + 1)); ++i)
    a[i] = (a[i] * (long long)a[i]) % MOD;
  FWT(a, 1 << (n + 1), 0);
  long long sum = 0;
  for (int i = 0; i < (1 << (n + 1)); i += 2) {
    int now = i >> 1;
    long long v = 1;
    for (int j = 0; j < (n); ++j) {
      if (!(now & 1)) v *= 2;
      now >>= 1;
    }
    sum = (sum + a[i] * v) % MOD;
  }
  cout << sum * 3 % MOD * INV_2 % MOD << endl;
  return 0;
}
