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 N = 20, P = 1000000007, INV2 = (P + 1) / 2;
using ll = long long;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % P;
    a = a * a % P;
    b >>= 1;
  }
  return ret;
}
ll c[1 << N];
char s[1 << N];
int n;
void FWT_xor(int fl) {
  for (int l = 1; l < n; l *= 2)
    for (int s = 0; s < n; s += l * 2)
      for (int k = s; k < s + l; k++) {
        ll tmp = c[k + l];
        c[k + l] = (c[k] + P - tmp) % P * (fl == 1 ? 1 : INV2) % P;
        c[k] = (c[k] + tmp) % P * (fl == 1 ? 1 : INV2) % P;
      }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> s;
  int nn = n;
  n = 1 << n;
  for (int i = 0; i < n; i++) c[i] = s[i] - '0';
  FWT_xor(1);
  for (int i = 0; i < n; i++) c[i] = c[i] * c[i] % P;
  FWT_xor(-1);
  ll ans = 0;
  for (int i = 0; i < n; i++)
    ans = (ans + c[i] * qpow(2, nn - __builtin_popcount(i)) % P) % P;
  cout << ans * 3 % P << endl;
  return 0;
}
