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 MAXN = (1 << 21) + 3;
const int ha = 1e9 + 7;
const int inv2 = 500000004;
char str[MAXN];
int n, N;
int pw[MAXN], pc[MAXN];
int f[MAXN];
inline void FWT(int f[]) {
  for (int mid = 1; mid < N; mid <<= 1) {
    for (int i = 0; i < N; i += (mid << 1)) {
      for (int j = 0; j <= mid - 1; ++j) {
        int x = f[i + j], y = f[i + mid + j];
        f[i + j] = (x + y) % ha;
        f[i + mid + j] = (x + ha - y) % ha;
      }
    }
  }
}
inline void iFWT(int f[]) {
  for (int mid = N >> 1; mid; mid >>= 1) {
    for (int i = 0; i < N; i += (mid << 1)) {
      for (int j = 0; j <= mid - 1; ++j) {
        int x = f[i + j], y = f[i + mid + j];
        f[i + j] = 1ll * (x + y) * inv2 % ha;
        f[i + mid + j] = 1ll * (x + ha - y) * inv2 % ha;
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  scanf("%s", str);
  N = (1 << (n + 1));
  pw[0] = 1;
  for (int i = 1; i <= MAXN - 1; ++i)
    pw[i] = 2ll * pw[i - 1] % ha, pc[i] = pc[i >> 1] + (i & 1);
  for (int i = 0; i <= (1 << n) - 1; ++i) f[i] = str[i] - '0';
  FWT(f);
  for (int i = 0; i <= N - 1; ++i) f[i] = 1ll * f[i] * f[i] % ha;
  iFWT(f);
  int ans = 0;
  for (int i = 0; i <= N - 1; ++i)
    (ans += 1ll * f[i] * pw[n - pc[i]] % ha) %= ha;
  ans = 3ll * ans % ha;
  printf("%d\n", ans);
  return 0;
}
