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>
const int maxn = 1048600;
int n, m;
char s[maxn];
int a[maxn], bits[maxn];
long long b[maxn], ans;
template <typename T>
inline void FWT(T *a) noexcept {
  T *End = a + n;
  for (register int hl = 1, l = 2; hl < n; hl = l, l <<= 1) {
    for (register T *i = a; i != End; i += l) {
      T *tmp = i + hl, *j = i, *k = tmp;
      for (; j != tmp; ++j, ++k) {
        T foo = *j;
        *j += *k;
        *k = foo - *k;
      }
    }
  }
  return;
}
int main(int argc, char const *argv[]) {
  scanf("%d", &m);
  n = 1 << m;
  fread(s, 1, maxn, stdin);
  char *S = s;
  while (*S != '0' && *S != '1') {
    ++S;
  }
  a[0] = *S - '0';
  ++S;
  for (register int i = 1; i < n; ++i) {
    a[i] = *S - '0';
    ++S;
    bits[i] = bits[i ^ (i & -i)] + 1;
  }
  FWT(a);
  for (register int i = 0; i < n; ++i) {
    b[i] = 1ll * a[i] * a[i];
  }
  FWT(b);
  for (register int i = 0; i < n; ++i) {
    ans += b[i] >> bits[i];
  }
  printf("%d\n", 3ll * ans % 1000000007);
  return 0;
}
