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 Mo = 1000000007;
using namespace std;
int i, j, m, n, p, k, inv2 = 500000004, A[1100005], B[1100005], ans, N;
char c[1100005];
int power(int x, int y) {
  int sum = 1;
  for (; y; y >>= 1) {
    if (y & 1) sum = 1ll * sum * x % Mo;
    x = 1ll * x * x % Mo;
  }
  return sum;
}
void FWT(int *a) {
  int i, j, k, l, t, x, y;
  for (i = 1; i < N; i <<= 1)
    for (j = 0, l = N / (i << 1); j < N; j += i << 1)
      for (k = 0, l = 0; k < i; ++k, l += t) {
        x = a[j + k];
        y = a[j + k + i];
        a[j + k] = (x + y) % Mo;
        a[j + k + i] = (x - y + Mo) % Mo;
      }
}
void NFWT(int *a) {
  int i, j, k, l, t, x, y;
  for (i = 1; i < N; i <<= 1)
    for (j = 0, l = N / (i << 1); j < N; j += i << 1)
      for (k = 0, l = 0; k < i; ++k, l += t) {
        x = a[j + k];
        y = a[j + k + i];
        a[j + k] = 1ll * (x + y) * inv2 % Mo;
        a[j + k + i] = 1ll * (x - y + Mo) * inv2 % Mo;
      }
}
int size(int x) {
  int s = 0;
  for (; x; x >>= 1) s += (x & 1);
  return s;
}
int main() {
  scanf("%d", &n);
  scanf("%s", c);
  for (i = 0; i < (1 << n); ++i) A[i] = c[i] - '0';
  for (i = 0; i < (1 << n); ++i) B[i] = 1 - A[i];
  N = 1 << n;
  FWT(A);
  FWT(B);
  for (i = 0; i < (1 << n); ++i) A[i] = 1ll * A[i] * B[i] % Mo;
  NFWT(A);
  for (i = 0; i < (1 << n); ++i)
    (ans += 3ll * A[i] * power(2, size(i)) % Mo) %= Mo;
  printf("%d\n", ans);
}
