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;
const int P = 1 << N;
const int mod = 1e9 + 7;
int power(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) {
      res = (1LL * res * a) % mod;
    }
    a = (1LL * a * a) % mod;
    b >>= 1;
  }
  return res;
}
int n;
char str[P + 1];
int a[P];
int p;
const int fst[2] = {0, mod};
int ans;
void go() {
  for (int i = 2; i <= p; i <<= 1) {
    int m = i >> 1;
    int u, v, x, y;
    for (int j = 0; j < p; j += i) {
      for (int k = 0; k < m; ++k) {
        u = a[j + k];
        v = a[j + k + m];
        x = u + v;
        y = u - v;
        a[j + k] = x - fst[x >= mod];
        a[j + k + m] = y + fst[y < 0];
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  scanf("%s", str);
  p = 1 << n;
  for (int i = 0; i < p; ++i) {
    a[i] = str[i] - '0';
  }
  go();
  for (int i = 0; i < p; ++i) {
    a[i] = (1LL * a[i] * a[i]) % mod;
  }
  go();
  int inv = power(p, mod - 2);
  for (int i = 0; i < p; ++i) {
    a[i] = (1LL * a[i] * inv) % mod;
  }
  for (int i = 0; i < p; ++i) {
    int tot = a[i];
    for (int j = 0; j < n; ++j) {
      if (!((i >> j) & 1)) {
        tot <<= 1;
        tot -= fst[tot >= mod];
      }
    }
    ans += tot;
    ans -= fst[ans >= mod];
  }
  ans = (ans * 3LL) % mod;
  printf("%d\n", ans);
}
