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 mod = 1e9 + 7;
int inverse(int x) {
  return x == 1 ? x : (long long)(mod - mod / x) * inverse(mod % x) % mod;
}
void FWT(vector<int> &a) {
  int n = a.size();
  for (int d = 1; d < n; d <<= 1)
    for (int m = d << 1, i = 0; i < n; i += m)
      for (int j = 0; j < d; j++) {
        int x = a[i + j], y = a[i + j + d];
        a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod;
      }
}
void IFWT(vector<int> &a) {
  int n = a.size();
  int inv2 = inverse(2);
  for (int d = 1; d < n; d <<= 1)
    for (int m = d << 1, i = 0; i < n; i += m)
      for (int j = 0; j < d; j++) {
        int x = a[i + j], y = a[i + j + d];
        a[i + j] = 1LL * (x + y) * inv2 % mod;
        a[i + j + d] = (1LL * (x - y) * inv2 % mod + mod) % mod;
      }
}
int main(int argc, char *argv[]) {
  int n;
  string s;
  cin >> n >> s;
  vector<int> a(1 << n);
  for (int i = 0; i < 1 << n; i++) {
    a[i] = s[i] - '0';
  }
  FWT(a);
  for (int i = 0; i < 1 << n; i++) {
    a[i] = (long long)a[i] * a[i] % mod;
  }
  IFWT(a);
  long long ans = 0;
  for (int i = 0; i < 1 << n; i++) {
    ans = (ans + (long long)(1 << (n - __builtin_popcount(i))) * a[i]) % mod;
  }
  cout << ans * 3 % mod << endl;
  return 0;
}
