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>
void read(T &x) {
  x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
}
const int N = 2200000;
const int P = 1e9 + 7;
const int inv2 = (P + 1) / 2;
long long mi[N], bl[N];
long long f[N], n, m;
char s[N];
inline void FWT_xor(long long *f, long long opt = 1) {
  for (int p = 2, k = 1; p <= m; p <<= 1, k <<= 1)
    for (int i = 0; i < m; i += p)
      for (int j = 0; j < k; j++) {
        long long x = f[i + j], y = f[i + j + k];
        f[i + j] = (x + y) % P;
        f[i + j + k] = (x + P - y) % P;
        if (opt == -1)
          f[i + j] = f[i + j] * inv2 % P,
                f[i + j + k] = f[i + j + k] * inv2 % P;
      }
}
int main() {
  read(n);
  m = 1 << n;
  scanf("%s", s);
  for (int i = 0; i < m; i++) f[i] = s[i] - '0';
  mi[0] = 1;
  for (int i = 1; i <= 30; i++) mi[i] = mi[i - 1] << 1;
  for (int i = 1; i <= m; i++) bl[i] = bl[i - (i & -i)] + 1;
  FWT_xor(f, 1);
  for (int i = 0; i < m; i++) f[i] = f[i] * f[i] % P;
  FWT_xor(f, -1);
  long long ans = f[0] * m % P;
  for (int i = 1; i < m; i++) ans = (ans + f[i] * mi[n - bl[i]]) % P;
  cout << ans * 3 % P << endl;
  return 0;
}
