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 mo = 1e9 + 7;
int n, a[1 << 20], c[1 << 20], ans;
long long fpm(long long x, long long y) {
  long long s = 1;
  while (y) {
    if (y & 1) s = (s * x) % mo;
    y >>= 1, x = (x * x) % mo;
  }
  return s;
}
int main() {
  int i, j, k, x, y;
  scanf("%d\n", &n), c[0] = (1 << n), ans = 0;
  for (i = 0; i < (1 << n); i++) a[i] = getchar() - '0';
  for (i = 1; i < (1 << n); i++) c[i] = (c[i - (i & (-i))] >> 1);
  for (i = 2; i <= (1 << n); i <<= 1)
    for (j = 0; j < (1 << n); j += i)
      for (k = 0; k < i / 2; k++) {
        x = a[j + k], y = a[j + k + i / 2];
        a[j + k] = (x + y >= mo ? x + y - mo : x + y);
        a[j + k + i / 2] = (x - y < 0 ? x - y + mo : x - y);
      }
  for (i = 0; i < (1 << n); i++) a[i] = (1LL * a[i] * a[i]) % mo;
  for (i = 2; i <= (1 << n); i <<= 1)
    for (j = 0; j < (1 << n); j += i)
      for (k = 0; k < i / 2; k++) {
        x = a[j + k], y = a[j + k + i / 2];
        a[j + k] = (x + y >= mo ? x + y - mo : x + y);
        a[j + k + i / 2] = (x - y < 0 ? x - y + mo : x - y);
      }
  x = fpm(1 << n, mo - 2);
  for (i = 0; i < (1 << n); i++) a[i] = (1LL * a[i] * x) % mo;
  for (i = 0; i < (1 << n); i++) ans = (ans + 1LL * a[i] * c[i]) % mo;
  ans = (3LL * ans) % mo;
  cout << ans;
  return 0;
}
