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>
const int Inv2 = (1000000007 + 1) >> 1;
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, num[1 << 20], s[1 << 20], ans;
void FWT(int* a, int l, int r, int v) {
  if (l == r) return;
  int mid = l + r >> 1;
  FWT(a, l, mid, v);
  FWT(a, mid + 1, r, v);
  for (int i = 0; i <= mid - l; ++i) {
    int x = a[l + i], y = a[mid + 1 + i];
    a[l + i] = 1LL * (x + y) * v % 1000000007;
    a[mid + 1 + i] = 1LL * (x - y + 1000000007) * v % 1000000007;
  }
}
int main() {
  n = read();
  for (int i = 0; i < 1 << n; ++i) s[i] = getchar() - '0';
  FWT(s, 0, (1 << n) - 1, 1);
  for (int i = 0; i < 1 << n; ++i) s[i] = 1LL * s[i] * s[i] % 1000000007;
  FWT(s, 0, (1 << n) - 1, Inv2);
  for (int i = 0; i < 1 << n; ++i) {
    num[i] = 1;
    for (int j = 0; j < n; ++j)
      if (!(i & (1 << j))) num[i] = 2 * num[i] % 1000000007;
    ans = (ans + 1LL * num[i] * s[i]) % 1000000007;
  }
  printf("%d\n", 3LL * ans % 1000000007);
  return 0;
}
