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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
long long num, n, ans;
long long f[1200010], cnt[1200010], pw[1200010];
char s[1200010];
void FWT(long long *a, int type) {
  for (int len = 1; len < n; len <<= 1) {
    for (int i = 0; i < n; i += len << 1) {
      for (int j = i; j < i + len; ++j) {
        long long x = a[j], y = a[j + len];
        a[j] = (x + y) % 1000000007,
        a[j + len] = (x - y + 1000000007) % 1000000007;
        if (type != 1)
          a[j] = a[j] * 500000004 % 1000000007,
          a[j + len] = a[j + len] * 500000004 % 1000000007;
      }
    }
  }
}
int main() {
  read(num), scanf("%s", s), pw[0] = 1;
  for (int i = 1; i <= num; ++i) pw[i] = pw[i - 1] * 2 % 1000000007;
  n = 1 << num;
  for (int i = 1; i < n; ++i) cnt[i] = cnt[i - (i & (-i))] + 1;
  for (int i = 0; i < n; ++i) f[i] = s[i] - '0';
  FWT(f, 1);
  for (int i = 0; i < n; ++i) f[i] = f[i] * f[i] % 1000000007;
  FWT(f, -1);
  for (int i = 0; i < n; ++i)
    ans = (ans + f[i] * pw[num - cnt[i]] % 1000000007) % 1000000007;
  printf("%lld", ans * 3 % 1000000007);
  return 0;
}
