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>
char s[(1 << 20) + 5];
int n, a[(1 << 20) + 5], b[(1 << 20) + 5];
inline int mod(int x) { return x < 1000000007 ? x : x - 1000000007; }
void fwt(long long v) {
  for (int i = 0; i < n; ++i)
    for (int j = 0, x; j < 1 << n; ++j)
      if (x = j ^ (1 << i), j & (1 << i))
        a[x] = mod(a[x] + a[j]),
        a[j] = v * mod(a[x] + 1000000007 - mod(a[j] << 1)) % 1000000007,
        a[x] = v * a[x] % 1000000007;
}
int main() {
  int i, j, ans = 0;
  scanf("%d%s", &n, s);
  for (i = 0; i < 1 << n; ++i) a[i] = s[i] - '0';
  fwt(1);
  for (i = 0; i < 1 << n; ++i) a[i] = 1LL * a[i] * a[i] % 1000000007;
  fwt((1000000007 + 1) >> 1);
  for (i = 0; i < 1 << n; ++i) {
    for (j = 0; j < n; ++j)
      if (~i & (1 << j)) a[i] = mod(a[i] << 1);
    ans = mod(ans + a[i]);
  }
  printf("%d", 3LL * ans % 1000000007);
}
