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;
int IN() {
  int c, f, x;
  while (!isdigit(c = getchar()) && c != '-')
    ;
  c == '-' ? (f = 1, x = 0) : (f = 0, x = c - '0');
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + c - '0';
  return !f ? x : -x;
}
const int N = 1 << 20;
const int p = 1e9 + 7;
int Pow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % p)
    if (b & 1) res = 1ll * res * a % p;
  return res;
}
int f[N], g[N];
int n, res;
char s[N + 19];
void FWT(int *A, int n, int f) {
  for (int i = 1; i < n; i <<= 1)
    for (int j = 0; j < n; j += i << 1) {
      for (int k = j; k < j + i; k++) {
        int u = A[k], v = A[k + i];
        A[k] = (u + v) % p;
        A[k + i] = (u - v + p) % p;
      }
    }
  if (f) {
    int tmp = Pow(n, p - 2);
    for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * tmp % p;
  }
}
int main() {
  n = IN();
  scanf("%s", s);
  for (int i = 0; i < 1 << n; i++) f[i] = s[i] - '0';
  FWT(f, 1 << n, 0);
  for (int i = 0; i < 1 << n; i++) f[i] = 1ll * f[i] * f[i] % p;
  FWT(f, 1 << n, 1);
  for (int t = 0; t < 1 << n; t++) {
    g[t] = 1 << n - __builtin_popcount(t);
    res = (res + 1ll * f[t] * g[t]) % p;
  }
  res = 3ll * res % p;
  printf("%d\n", res);
}
