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;
char getc() {
  char c = getchar();
  while ((c < 'A' || c > 'Z') && (c < 'a' || c > 'z') && (c < '0' || c > '9'))
    c = getchar();
  return c;
}
int gcd(int n, int m) { return m == 0 ? n : gcd(m, n % m); }
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return x * f;
}
int n, a[1 << 21], p[21], ans;
char s[1 << 21];
void FWT(int n, int *a, int op) {
  for (int i = 2; i <= n; i <<= 1)
    for (int j = 0; j < n; j += i)
      for (int k = j; k < j + (i >> 1); k++) {
        int x = a[k], y = a[k + (i >> 1)];
        a[k] = (x + y) % 1000000007,
        a[k + (i >> 1)] = (x - y + 1000000007) % 1000000007;
        if (op == 1)
          a[k] = 1ll * a[k] * 500000004 % 1000000007,
          a[k + (i >> 1)] = 1ll * a[k + (i >> 1)] * 500000004 % 1000000007;
      }
}
signed main() {
  n = read();
  scanf("%s", s);
  for (int i = 0; i < (1 << n); i++) a[i] = s[i] - '0';
  FWT(1 << n, a, 0);
  for (int i = 0; i < (1 << n); i++) a[i] = 1ll * a[i] * a[i] % 1000000007;
  FWT(1 << n, a, 1);
  for (int i = 0; i < (1 << n); i++) {
    int u = n, j = i;
    while (j) j ^= j & -j, u--;
    ans = (ans + 1ll * a[i] * (1 << u)) % 1000000007;
  }
  cout << 3ll * ans % 1000000007;
  return 0;
}
