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;
inline int M(int x) { return (x >= 1000000007) ? (x - 1000000007) : x; }
inline int M1(int x) { return (x < 0) ? (x + 1000000007) : x; }
inline int ksm(int p, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * p % 1000000007;
    k >>= 1;
    p = 1ll * p * p % 1000000007;
  }
  return res;
}
int n, all, siz[1048577], f[1048577], ans = 0;
char sc[1048577];
inline void FWT(int *f, int len) {
  int i, j, x, y;
  for (j = 1; j < len; j <<= 1)
    for (i = 0; i < len; i++)
      if (i & j) x = f[i ^ j], y = f[i], f[i ^ j] = M(x + y), f[i] = M1(x - y);
}
inline void IFWT(int *f, int len) {
  int i, j, x, y, inv = ksm(len, 1000000007 - 2);
  for (j = 1; j < len; j <<= 1)
    for (i = 0; i < len; i++)
      if (i & j) x = f[i ^ j], y = f[i], f[i ^ j] = M(x + y), f[i] = M1(x - y);
  for (i = 0; i < len; i++) f[i] = 1ll * f[i] * inv % 1000000007;
}
int main() {
  for (int i = 1; i < 1048577; i++) siz[i] = siz[i >> 1] + (i & 1);
  scanf("%d", &n);
  all = 1 << n;
  int i;
  scanf("%s", sc);
  for (i = 0; i < all; i++) f[i] = sc[i] ^ 48;
  FWT(f, all);
  for (i = 0; i < all; i++)
    f[i] = 1ll * f[i] * f[i] % 1000000007 *
           ((siz[i] & 1) ? (1000000007 - 1) : 1) % 1000000007;
  IFWT(f, all);
  for (i = 0; i < all; i++)
    ans = (ans + 1ll * f[i] * (1 << siz[i])) % 1000000007;
  ans = 3ll * ans % 1000000007;
  printf("%d", ans);
  return 0;
}
