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 Mod = 1e9 + 7;
int n, U;
long long a[1 << N], b[1 << N], c[1 << N];
char ss[(1 << N) + 10];
void trans(long long a[], int flag) {
  for (int b = 0; b < n; b++) {
    int u = U ^ (1 << b);
    for (int s = u, t = 1 << (n - 1); t; s = (s - 1) & u, t--) {
      long long l = a[s], r = a[s | (1 << b)];
      if (flag == 1) {
        a[s] = l + r;
        a[s | (1 << b)] = l - r;
      } else {
        a[s] = (l + r) / 2;
        a[s | (1 << b)] = (l - r) / 2;
      }
    }
  }
}
int bitscount(int a) {
  int rt = 0;
  while (a) {
    rt++;
    a -= a & -a;
  }
  return rt;
}
int main() {
  scanf("%d", &n);
  U = (1 << n) - 1;
  scanf("%s", ss);
  for (int i = 0; i <= U; i++) a[i] = ss[i] - '0';
  for (int i = 0; i <= U; i++) b[i] = 1 - (ss[i] - '0');
  trans(a, 1);
  trans(b, 1);
  for (int s = 0; s <= U; s++) c[s] = a[s] * b[s];
  trans(c, -1);
  long long ans = 0;
  for (int i = 0; i <= U; i++) {
    ans += 1LL * c[i] * (1LL << bitscount(i));
  }
  printf("%d\n", (int)(ans * 3 % Mod));
}
