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;
char s[(1 << 20) + 1];
void FWHT(vector<long long>& P, bool inverse) {
  int n = 1;
  for (; n < P.size(); n = (n << 1))
    ;
  P.resize(n);
  for (int len = 1; len < n; len = (len << 1)) {
    for (int i = 0; i < n; i += (len + len)) {
      for (int j = 0; j < len; ++j) {
        long long u = P[i + j], v = P[i + len + j];
        P[i + j] = u + v;
        P[i + len + j] = u - v;
      }
    }
  }
  if (inverse) {
    for (int i = 0; i < n; ++i) P[i] /= n;
  }
}
int main() {
  int n;
  vector<long long> p;
  long long ans = 0;
  scanf("%d%s", &n, s);
  p.resize((1 << n));
  for (int i = 0; i < (1 << n); ++i) p[i] = (s[i] - '0');
  FWHT(p, 0);
  for (int i = 0; i < (1 << n); ++i) p[i] *= p[i];
  FWHT(p, 1);
  for (int i = 0; i < (1 << n); ++i)
    ans = (ans + p[i] * (1 << (n - __builtin_popcount(i)))) % 1000000007;
  ans = (3 * ans) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}
