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 long long mod = 1000000007;
long long ans, n, f[3000000];
char s[3000000];
long long ppc(long long x) {
  long long ret = 0;
  for (; x; x >>= 1) ret += x & 1;
  return ret;
}
long long inv(long long x) {
  return x == 1 ? 1 : (mod - mod / x) * inv(mod % x) % mod;
}
void fwt(long long f[], long long n) {
  for (long long i = 0; i < n; ++i) {
    long long t = 1 << i;
    for (long long j = 0; j < 1 << n; j += t << 1) {
      long long *f1 = f + j, *f2 = f + j + t;
      for (long long k = 0; k < t; ++k) {
        long long x = f1[k], y = f2[k];
        f1[k] = (x + y) % mod;
        f2[k] = (x - y + mod) % mod;
      }
    }
  }
}
int main() {
  scanf("%lld%s", &n, s);
  for (long long i = 0; i < 1 << n; ++i) f[i] = s[i] - '0';
  fwt(f, n);
  for (long long i = 0; i < 1 << n; ++i) f[i] = f[i] * f[i] % mod;
  fwt(f, n);
  for (long long i = 0, ii = inv(1 << n); i < 1 << n; ++i) {
    f[i] = f[i] * ii % mod;
    ans = (ans + f[i] * (1 << (n - ppc(i)))) % mod;
  }
  printf("%lld\n", ans * 3 % mod);
  return 0;
}
