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;
const int N = 1 << 20;
const int mod = 1e9 + 7, inv2 = 500000004;
template <class T>
inline void Inc(T &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
template <class T>
inline void Dec(T &x, int y) {
  x -= y;
  if (x < 0) x += mod;
}
template <class T>
inline void Mod(T &x) {
  if (x >= mod)
    x -= mod;
  else if (x < 0)
    x += mod;
}
char S[N];
int n;
int f[N], g[N], cnt[N];
int len;
inline void FWT(int *A, int n, int f) {
  for (int i = 1; i < n; i <<= 1) {
    for (int p = i << 1, j = 0; j < n; j += p) {
      for (int k = 0; k < i; ++k) {
        int X = A[j | k], Y = A[j | k | i];
        Inc(A[j | k], Y);
        A[j | k | i] = X, Dec(A[j | k | i], Y);
        if (f == -1)
          A[j | k] = (long long)inv2 * A[j | k] % mod,
                A[j | k | i] = (long long)inv2 * A[j | k | i] % mod;
      }
    }
  }
  return;
}
int main() {
  cin >> len;
  n = 1 << len;
  cin >> S;
  for (int i = 0; i < n; ++i) {
    f[i] = S[i] - '0', g[i] = f[i] ^ 1;
    cnt[i] = cnt[i >> 1] + (i & 1);
  }
  FWT(f, n, 1);
  FWT(g, n, 1);
  for (int i = 0; i < n; ++i) f[i] = (long long)f[i] * g[i] % mod;
  FWT(f, n, -1);
  int ans = 0;
  for (int i = 0; i < n; ++i) Inc(ans, (long long)(1 << cnt[i]) * f[i] % mod);
  ans = (long long)ans * 3 % mod;
  cout << ans << endl;
}
