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 long long mo = 1e9 + 7, inv2 = 5e8 + 4;
int n, N;
string st;
long long A[1 << 20];
long long C[1 << 20];
void get_read() {
  ios::sync_with_stdio(false);
  cin >> n >> st;
}
void FWT(long long *P, int opt) {
  for (int i = 2; i <= N; i <<= 1)
    for (int p = i >> 1, j = 0; j < N; j += i)
      for (int k = j; k < j + p; ++k) {
        long long X = P[k], Y = P[k + p];
        P[k] = (X + Y) % mo;
        P[k + p] = ((X - Y) % mo + mo) % mo;
        if (opt == -1) P[k] = P[k] * inv2 % mo, P[k + p] = P[k + p] * inv2 % mo;
      }
}
long long ksm(long long a, long long b) {
  long long s = 1;
  while (b) {
    if (b & 1) s = s * a % mo;
    a = a * a % mo;
    b >>= 1;
  }
  return s;
}
void get_work() {
  N = (1 << n);
  for (int i = 0; i < N; ++i) A[i] = st[i] - '0';
  FWT(A, 1);
  for (int i = 0; i < N; ++i) A[i] = A[i] * A[i] % mo;
  FWT(A, -1);
  long long ans = 0;
  for (int i = 0; i < N; ++i) {
    C[i] = C[i >> 1] + (i & 1);
    ans = (ans + A[i] * ksm(2, n - C[i])) % mo;
  }
  printf("%lld", (3 * ans % mo + mo) % mo);
}
int main() {
  get_read();
  get_work();
  return 0;
}
