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 int mod = 1000000007;
long long p2[30];
void FST(vector<long long>& a, bool inv) {
  for (int n = (int)(a).size(), step = 1; step < n; step *= 2) {
    for (int i = 0; i < n; i += 2 * step)
      for (int j = i; j < (i + step); ++j) {
        long long &u = a[j], &v = a[j + step];
        tie(u, v) = pair<long long, long long>(u + v, u - v);
      }
  }
  if (inv)
    for (long long& x : a) x /= (int)(a).size();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  p2[0] = 1;
  for (int i = 1; i <= 25; i++) p2[i] = (p2[i - 1] + p2[i - 1]) % mod;
  int n;
  string s;
  cin >> n >> s;
  vector<long long> a((int)(s).size());
  for (int i = 0; i < (1 << n); i++) {
    a[i] = s[i] - '0';
  }
  FST(a, false);
  for (int i = 0; i < (int)(a).size(); i++) {
    a[i] = (a[i] * 1ll * a[i]);
  }
  FST(a, true);
  long long ans = 0;
  for (int i = 0; i < (int)(a).size(); i++) {
    ans = (ans + a[i] * p2[n - __builtin_popcount(i)]) % mod;
  }
  ans = (ans * 3) % mod;
  cout << ans << endl;
  return 0;
}
