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 long long Mo = 1000000007LL;
int N, n;
long long A[1 << 20], B[1 << 20], C[1 << 20], js[1 << 20];
char s[1 << 21];
void fwt_xor(long long *A, int n) {
  if (n == 1) return;
  fwt_xor(A, n >> 1);
  fwt_xor(A + (n >> 1), n >> 1);
  for (int i = 0; i < n; i++) js[i] = A[i];
  for (int i = 0; i < n >> 1; i++) A[i] = js[i] + js[i + (n >> 1)];
  for (int i = (n >> 1); i < n; i++) A[i] = js[i - (n >> 1)] - js[i];
  return;
}
void ufwt_xor(long long *A, int n) {
  if (n == 1) return;
  for (int i = 0; i < n; i++) js[i] = A[i];
  for (int i = 0; i < n >> 1; i++) A[i] = (js[i] + js[i + (n >> 1)]) / 2;
  for (int i = (n >> 1); i < n; i++) A[i] = (js[i - (n >> 1)] - js[i]) / 2;
  ufwt_xor(A, n >> 1);
  ufwt_xor(A + (n >> 1), n >> 1);
  return;
}
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  N = 1 << n;
  for (int i = 0; i < N; i++) A[i] = B[i] = s[i] - '0';
  fwt_xor(A, N);
  fwt_xor(B, N);
  for (int i = 0; i < N; i++) C[i] = A[i] * B[i];
  ufwt_xor(C, N);
  long long ans = 0LL;
  for (int st = 0; st < N; st++) {
    long long sing = C[st] % Mo;
    for (int i = 0; i < n; i++)
      if (!((st >> i) & 1)) sing = (sing * 2LL) % Mo;
    ans += sing;
    ans %= Mo;
  }
  cout << ans * 3LL % Mo << endl;
  return 0;
}
