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 maxn(20);
const int maxm(1 << 20);
const int mod(1e9 + 7);
const int inv2((mod + 1) >> 1);
inline void Inc(int &x, int y) { x = x + y >= mod ? x + y - mod : x + y; }
inline void Dec(int &x, int y) { x = x - y < 0 ? x - y + mod : x - y; }
inline int Add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int Sub(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline int Pow(long long x, int y) {
  long long ret = 1;
  for (; y; y >>= 1, x = x * x % mod)
    if (y & 1) ret = ret * x % mod;
  return ret;
}
int n, s, f[maxm], bit[maxm], ans;
inline void FWT(int *p, int opt) {
  int i, j, k, t, x, y;
  for (i = 1; i < s; i <<= 1)
    for (j = 0, t = i << 1; j < s; j += t)
      for (k = 0; k < i; ++k) {
        x = p[j + k], y = p[j + k + i];
        p[j + k] = Add(x, y), p[j + k + i] = Sub(x, y);
        if (opt == -1)
          p[j + k] = (long long)p[j + k] * inv2 % mod,
                p[j + k + i] = (long long)p[j + k + i] * inv2 % mod;
      }
}
int main() {
  int i;
  scanf("%d", &n), s = 1 << n;
  for (i = 0; i < s; ++i) {
    scanf(" %c", &f[i]), f[i] -= '0';
    bit[i] = bit[i >> 1] + (i & 1);
  }
  FWT(f, 1);
  for (i = 0; i < s; ++i) f[i] = (long long)f[i] * f[i] % mod;
  FWT(f, -1);
  for (i = 0; i < s; ++i) Inc(ans, (long long)(1 << (n - bit[i])) * f[i] % mod);
  ans = (long long)ans * 3 % mod, printf("%d\n", ans);
  return 0;
}
