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;
inline int gi() {
  int data = 0, m = 1;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') {
    m = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    data = (data << 1) + (data << 3) + (ch ^ 48);
    ch = getchar();
  }
  return (m) ? data : -data;
}
int n, Mx, f1[1048576], f2[1048576], cnt[1048576], Ans;
char s[1048576];
inline void FWT(int *P, int op) {
  for (int i = 1; i < Mx; i <<= 1)
    for (int j = 0, dt = (i << 1); j < Mx; j += dt)
      for (int k = 0; k < i; k++) {
        int X = P[j + k], Y = P[j + k + i];
        P[j + k] = 1ll * (X + Y) * op % 1000000007;
        P[j + k + i] = 1ll * (X - Y + 1000000007) * op % 1000000007;
      }
  return;
}
int main() {
  n = gi();
  Mx = (1 << n);
  for (int i = 0; i < Mx; i++) scanf("%s", s);
  for (int i = 0; i < Mx; i++)
    if (s[i] == '1') f1[i] = 1;
  for (int i = 0; i < Mx; i++)
    if (s[i] == '1') f2[i] = 1;
  FWT(f1, 1);
  FWT(f2, 1);
  for (int i = 0; i < Mx; i++) f1[i] = 1ll * f1[i] * f2[i] % 1000000007;
  FWT(f1, (1000000007 + 1) >> 1);
  for (int i = 0; i < Mx; i++) cnt[i] = cnt[i >> 1] + (i & 1);
  Ans = 0;
  for (int i = 0; i < Mx; i++)
    Ans = (Ans + 1ll * f1[i] * (1 << (n - cnt[i])) % 1000000007) % 1000000007;
  cout << 3ll * Ans % 1000000007 << endl;
  return 0;
}
