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;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x < y ? x : y;
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
const int N = 1 << 20, Mod = 1e9 + 7, inv = 5e8 + 4;
int f[N], g[N], sum[N];
void fwt(int *a, int n, int fl) {
  for (int i = 1; i < n; i <<= 1)
    for (int j = 0; j < n; j += i << 1)
      for (int k = 0; k < i; k++) {
        int tx = a[j + k], ty = a[i + j + k];
        a[j + k] = (tx + ty) % Mod, a[i + j + k] = (tx - ty + Mod) % Mod;
        if (fl == -1)
          a[j + k] = 1ll * a[j + k] * inv % Mod,
                a[i + j + k] = 1ll * a[i + j + k] * inv % Mod;
      }
}
int main() {
  int n;
  read(n);
  int msk = 1 << n;
  for (int i = 0; i < msk; i++) {
    char c = getchar();
    while (!isdigit(c)) c = getchar();
    f[i] = c - '0', g[i] = f[i] ^ 1;
  }
  fwt(f, msk, 1), fwt(g, msk, 1);
  for (int i = 0; i < msk; i++) f[i] = 1ll * f[i] * g[i] % Mod;
  fwt(f, msk, -1);
  int ans = 0;
  for (int i = 1; i < msk; i++) sum[i] = sum[i >> 1] + (i & 1);
  for (int i = 0; i < msk; i++)
    ans = (ans + 1ll * f[i] * (1 << sum[i]) % Mod) % Mod;
  cout << 3ll * ans % Mod << '\n';
  return 0;
}
