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 mod = 1e9 + 7, inv2 = (mod + 1) >> 1;
inline int& reduce(int& x) { return x >= mod ? x -= mod : x; }
void fwt(int* f, int n, bool flag) {
  for (int p = 2, q = 1; p <= (1 << n); q = (p <<= 1) >> 1)
    for (int i = 0; i < (1 << n); i += p)
      for (int x, y, j = i; j < i + q; j++) {
        x = f[j], y = f[j + q], reduce(f[j] = x + y),
        reduce(f[j + q] = x - y + mod);
        if (flag)
          f[j] = (f[j] * 1ll * inv2) % mod,
          f[j + q] = (f[j + q] * 1ll * inv2) % mod;
      }
}
const int N = 20;
int n, f[1 << N];
signed main() {
  static char s[1 << N];
  scanf("%d%s", &n, s);
  for (int i = 0; i < (1 << n); i++) f[i] = s[i] - '0';
  fwt(f, n, 0);
  for (int i = 0; i < (1 << n); i++) f[i] = (f[i] * 1ll * f[i]) % mod;
  fwt(f, n, 1);
  int ans = 0;
  for (int i = 0; i < (1 << n); i++)
    reduce(ans += (f[i] * 1ll * (1 << (n - __builtin_popcount(i)))) % mod);
  return printf("%lld\n", ans * 3ll % mod), 0;
}
