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>
const int maxM = 21, maxN = 1 << maxM, p = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  return x >= p ? x - p : x;
}
inline int sub(int x, int y) {
  x -= y;
  return x >= 0 ? x : x + p;
}
inline int dv2(int x) { return x & 1 ? (x + p) >> 1 : x >> 1; }
int m, n, ans, a[maxN], g[maxM];
char str[maxN];
void fwt(int f[], int sgn) {
  for (int len = 1; len < n; len <<= 1) {
    for (int i = 0; i < n; i += len << 1) {
      for (int j = 0; j < len; j++) {
        int x = f[i + j], y = f[i + j + len];
        f[i + j] = sgn == 1 ? add(x, y) : dv2(add(x, y));
        f[i + j + len] = sgn == 1 ? sub(x, y) : dv2(sub(x, y));
      }
    }
  }
}
int main() {
  scanf("%d%s", &m, str);
  n = 1 << m;
  for (int i = g[0] = 1; i <= m; i++) {
    g[i] = g[i - 1] * 2 % p;
  }
  for (int i = 0; i < n; i++) {
    a[i] = str[i] ^ 49;
  }
  fwt(a, 1);
  for (int i = 0; i < n; i++) {
    a[i] = (long long)a[i] * a[i] % p;
  }
  fwt(a, -1);
  for (int i = 0; i < n; i++) {
    ans = add(ans, (long long)a[i] * g[m - __builtin_popcount(i)] % p);
  }
  printf("%d\n", (long long)ans * 3 % p);
  return 0;
}
