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 mod = 1e9 + 7;
const int MN = 1 << 20;
void fwht(long long *a, bool invert) {
  for (int len = 2; len <= MN; len <<= 1) {
    for (int i = 0; i < MN; i += len) {
      for (int j = 0; j < len / 2; j++) {
        long long u = a[i + j], v = a[i + j + len / 2];
        a[i + j] = u + v;
        a[i + j + len / 2] = u - v;
      }
    }
  }
  if (invert)
    for (int i = 0; i < MN; i++) a[i] /= MN;
}
long long fa[MN], fb[MN];
void multiply(long long *a, long long *b, long long *res) {
  for (int i = 0; i < MN; i++) fa[i] = a[i];
  for (int i = 0; i < MN; i++) fb[i] = b[i];
  fwht(fa, false);
  fwht(fb, false);
  for (int i = 0; i < MN; i++) fa[i] *= fb[i];
  fwht(fa, true);
  for (int i = 0; i < MN; i++) res[i] = fa[i];
}
int N;
long long A[MN], B[MN];
int main() {
  scanf("%d", &N);
  scanf("\n");
  for (int i = 0; i < (1 << N); i++) {
    char t;
    scanf("%c", &t);
    A[i] = t == '1';
  }
  multiply(A, A, B);
  int ans = 0;
  for (int i = 0; i < (1 << N); i++) {
    int p = 1;
    for (int j = 0; j < N; j++) {
      if (i & (1 << j)) continue;
      p = p * 2 % mod;
    }
    ans += 1LL * p * B[i] % mod;
    ans %= mod;
  }
  ans = 1LL * ans * 3 % mod;
  printf("%d", ans);
}
