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>
const int M = 1100005;
const int MOD = 1e9 + 7;
int read() {
  int num = 0, flag = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag = -1;
  while (c >= '0' && c <= '9')
    num = (num << 3) + (num << 1) + (c ^ 48), c = getchar();
  return num * flag;
}
int n, m, ans, f[M], bit[M], inv2 = (MOD + 1) / 2;
void fwt(int *a, int n, int op) {
  for (int i = 1; i < n; i <<= 1)
    for (int p = i << 1, j = 0; j < n; j += p)
      for (int k = 0; k < i; k++) {
        int x = a[j + k], y = a[i + j + k];
        a[j + k] = (x + y) % MOD;
        a[i + j + k] = (x + MOD - y) % MOD;
        if (op == -1) {
          a[j + k] = 1ll * a[j + k] * inv2 % MOD;
          a[i + j + k] = 1ll * a[i + j + k] * inv2 % MOD;
        }
      }
}
signed main() {
  n = read();
  m = 1 << n;
  for (int i = 0; i < m; i++) scanf("%1d", &f[i]);
  for (int i = 1; i < m; i++) bit[i] = bit[i >> 1] + (i & 1);
  fwt(f, m, 1);
  for (int i = 0; i < m; i++) f[i] = 1ll * f[i] * f[i] % MOD;
  fwt(f, m, -1);
  for (int i = 0; i < m; i++)
    ans = (ans + 1ll * (1 << (n - bit[i])) * f[i]) % MOD;
  printf("%d\n", ans * 3ll % MOD);
}
