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 M = 1e9 + 7;
const int N = 22;
long long a[1 << N];
long long powa(long long a, long long b) {
  a %= M;
  long long res = 1 % M;
  while (b > 0) {
    if (b & 1) res = res * a % M;
    b >>= 1;
    a = a * a % M;
  }
  return res;
}
void fwht(long long *data, int dim) {
  for (int len = 1; 2 * len <= dim; len <<= 1) {
    for (int i = 0; i < dim; i += 2 * len) {
      for (int j = 0; j < len; j++) {
        long long a = data[i + j];
        long long b = data[i + j + len];
        data[i + j] = (a + b) % M;
        data[i + j + len] = (M + a - b) % M;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(15);
  cout.setf(ios::fixed);
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < (1 << n); i++) {
    a[i] = s[i] - '0';
  }
  fwht(a, 1 << n);
  for (int i = 0; i < (1 << n); i++) a[i] = (1LL * a[i] * a[i]) % M;
  fwht(a, 1 << n);
  long long im = powa(1 << n, M - 2);
  for (int i = 0; i < (1 << n); i++) a[i] = (a[i] * im) % M;
  long long ans = 0;
  int pw[1 << n];
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = (pw[i - 1] + pw[i - 1]) % M;
  for (int i = 0; i < (1 << n); i++)
    ans = (ans + 1LL * a[i] * pw[n - __builtin_popcount(i)]) % M;
  cout << (3 * ans) % M << '\n';
}
