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;
const int N = 1 << 20, P = 1000000007, inv_2 = 500000004;
inline int add(int a, int b) { return a + b >= P ? a + b - P : a + b; }
inline int sub(int a, int b) { return a - b < 0 ? a - b + P : a - b; }
inline int mul(int a, int b) { return 1ll * a * b - 1ll * a * b / P * P; }
int f[N], cnt[N];
char s[N];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
inline void FWT(int *a, int n, int t) {
  for (register int i = 1; i < n; i <<= 1)
    for (register int j = 0; j < n; j += (i << 1))
      for (register int k = 0; k < i; ++k) {
        int x = a[j + k], y = a[i + j + k];
        a[j + k] = add(x, y);
        a[i + j + k] = sub(x, y);
        if (!~t) {
          a[j + k] = mul(a[j + k], inv_2);
          a[i + j + k] = mul(a[i + j + k], inv_2);
        }
      }
}
int main() {
  int n = read(), len = 1 << n;
  scanf("%s", s);
  for (int i = 0; i < len; ++i) f[i] = s[i] - '0';
  FWT(f, len, 1);
  for (int i = 0; i < len; ++i) f[i] = mul(f[i], f[i]);
  FWT(f, len, -1);
  for (int i = 0; i < len; ++i) cnt[i] = cnt[i >> 1] + (i & 1);
  int ans = 0;
  for (int i = 0; i < len; ++i) ans = add(ans, mul(f[i], (1 << (n - cnt[i]))));
  printf("%d\n", mul(ans, 3));
  return 0;
}
