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, P = 1e9 + 7, M = (1 << N) + 50, I = P - P / 2;
int mul(int a, int b) { return 1ll * a * b % P; }
int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + P : a;
}
int add(int a, int b) {
  a += b;
  return a >= P ? a - P : a;
}
void fwt(int n, int *x, int op) {
  for (int s = 2; s <= n; s <<= 1) {
    int m = s >> 1;
    for (int j = 0; j < n; j += s)
      for (int k = 0; k < (m); ++k) {
        int a = x[j + k], b = x[j + m + k];
        x[j + k] = add(a, b);
        x[j + m + k] = sub(a, b);
        if (op == -1) {
          x[j + k] = mul(x[j + k], I);
          x[j + m + k] = mul(x[j + m + k], I);
        }
      }
  }
}
void mul(int n, int *srca, int *srcb, int *dest) {
  static int a[M], b[M];
  for (int i = 0; i < (n); ++i) a[i] = srca[i], b[i] = srcb[i];
  fwt(n, a, 1), fwt(n, b, 1);
  for (int i = 0; i < (n); ++i) dest[i] = mul(a[i], b[i]);
  fwt(n, dest, -1);
}
int n, m, f[1 << N], g[1 << N], ans, cnt[1 << N], pwr[N + 1];
int main(int argc, char *argv[]) {
  scanf("%d", &n);
  m = 1 << n;
  for (int i = 0; i < (m); ++i) {
    char c;
    scanf(" %c", &c);
    f[i] = c - '0';
  }
  mul(m, f, f, g);
  cnt[0] = 0;
  for (int i = 1; i < (m); ++i) cnt[i] = cnt[i >> 1] + (i & 1);
  pwr[0] = 1;
  for (int i = 1; i <= (n); ++i) pwr[i] = mul(pwr[i - 1], 2);
  for (int i = 0; i < (m); ++i) {
    ans = add(ans, mul(g[i], pwr[n - cnt[i]]));
  }
  ans = mul(3, ans);
  printf("%d\n", ans);
  return 0;
}
