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 M = 1000000007;
int n, q, k;
int N, ans[1100005], cnt;
char c[1100005];
int qpow(int a, int b) {
  int s = a, ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * s % M;
    s = 1ll * s * s % M;
    b >>= 1;
  }
  return ans;
}
int inv2 = qpow(2, M - 2);
void FWT_xor(int *a, int opt) {
  for (register int i = 1; i < N; i <<= 1)
    for (register int p = i << 1, j = 0; j < N; j += p)
      for (register int k = 0; k < i; ++k) {
        int X = a[j + k], Y = a[i + j + k];
        a[j + k] = (X + Y) % M;
        a[i + j + k] = (X + M - Y) % M;
        if (opt == -1)
          a[j + k] = 1ll * a[j + k] * inv2 % M,
                a[i + j + k] = 1ll * a[i + j + k] * inv2 % M;
      }
}
int lowbit(int x) { return -x & x; }
int main() {
  scanf("%d", &n);
  N = 1 << n;
  scanf("%s", c);
  for (register int i = 0; i < N; ++i) ans[i] = c[i] - '0';
  FWT_xor(ans, 1);
  for (register int i = 0; i < N; ++i) ans[i] = 1ll * ans[i] * ans[i] % M;
  FWT_xor(ans, -1);
  for (register int i = 0; i < N; ++i) {
    int tmp = i, s = 0;
    while (tmp) {
      ++s;
      tmp -= lowbit(tmp);
    }
    cnt = (cnt + 1ll * qpow(2, n - s) * ans[i]) % M;
  }
  cout << 1ll * cnt * 3 % M;
}
