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 maxd = 1000000007, N = 100000, inv2 = 500000004;
const double pi = acos(-1.0);
int n, cnt[1100000];
long long bin[1100000], a[1100000], b[1100000];
char s[1100000];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while ((ch < '0') || (ch > '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ((ch >= '0') && (ch <= '9')) {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
void fwt(int lim, long long *a, int typ) {
  int mid;
  for (mid = 1; mid < lim; mid <<= 1) {
    int len = (mid << 1), sta, j;
    for (sta = 0; sta < lim; sta += len) {
      for (j = 0; j < mid; j++) {
        int x = a[sta + j], y = a[sta + j + mid];
        a[sta + j] = (x + y) % maxd;
        a[sta + j + mid] = (x - y + maxd) % maxd;
        if (typ == -1) {
          a[sta + j] = (a[sta + j] * inv2) % maxd;
          a[sta + j + mid] = (a[sta + j + mid] * inv2) % maxd;
        }
      }
    }
  }
}
int main() {
  n = read();
  int lim = (1 << n), i;
  scanf("%s", s);
  bin[0] = 1;
  for (i = 0; i < lim; i++) {
    a[i] = s[i] - '0';
    b[i] = a[i];
  }
  for (i = 1; i < lim; i++) {
    bin[i] = (bin[i - 1] << 1) % maxd;
    cnt[i] = cnt[i >> 1] + (i & 1);
  }
  fwt(lim, a, 1);
  fwt(lim, b, 1);
  for (i = 0; i < lim; i++) a[i] = (a[i] * b[i]) % maxd;
  fwt(lim, a, -1);
  long long ans = 0;
  for (i = 0; i < lim; i++)
    ans = ((a[i] * bin[n - cnt[i]]) % maxd + ans) % maxd;
  ans = (ans * 3) % maxd;
  printf("%lld", ans);
  return 0;
}
