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 MAXN = 20;
const int MOD = 1e9 + 7;
int nowarn;
int n;
long long win[1 << MAXN], val[1 << MAXN];
void xor_fwt(long long *a, int n) {
  for (int m = 2; m <= n; m <<= 1)
    for (int i = 0; i < n; i += m)
      for (int j = 0; j < (m >> 1); ++j) {
        long long x = a[i + j], y = a[i + j + (m >> 1)];
        a[i + j] = x + y;
        a[i + j + (m >> 1)] = x - y;
      }
}
void xor_ifwt(long long *a, int n) {
  for (int m = 2; m <= n; m <<= 1)
    for (int i = 0; i < n; i += m)
      for (int j = 0; j < (m >> 1); ++j) {
        long long x = a[i + j], y = a[i + j + (m >> 1)];
        a[i + j] = (x + y) >> 1;
        a[i + j + (m >> 1)] = (x - y) >> 1;
      }
}
int main() {
  nowarn = scanf("%d", &n);
  for (int i = 0; i < (1 << n); ++i) {
    char ch;
    nowarn = scanf(" %c", &ch);
    win[i] = (ch == '1');
    val[i] = 1 << n;
    for (int j = 0; j < n; ++j)
      if (i & (1 << j)) val[i] >>= 1;
  }
  xor_fwt(win, 1 << n);
  for (int i = 0; i < (1 << n); ++i) win[i] *= win[i];
  xor_ifwt(win, 1 << n);
  int ans = 0;
  for (int i = 0; i < (1 << n); ++i)
    ans = int((ans + win[i] * val[i] % MOD) % MOD);
  ans = int((long long)ans * 3 % MOD);
  printf("%d\n", ans);
  return 0;
}
