#include <bits/stdc++.h>
const int mo = 1e9 + 7, N = 1111111, inv2 = 5e8 + 4;
int n, all, v[N], a[N], cnt[N], mi[22], res;
char t[N];
int main() {
  scanf("%d", &n);
  all = (1 << n) - 1;
  scanf("%s", t);
  for (register int i = 0; i <= all; i++) v[i] = t[i] - '0';
  for (int j = 2; j <= (1 << n); j <<= 1)
    for (int i = 0; i < (1 << n); i += j)
      for (register int k = 0; k <= (j >> 1) - 1; k++) {
        int x = v[i + k], y = v[i + k + (j >> 1)];
        v[i + k] = x + y < mo ? x + y : x + y - mo;
        v[i + k + (j >> 1)] = x - y >= 0 ? x - y : x - y + mo;
      }
  for (register int i = 0; i <= all; i++) a[i] = 1LL * v[i] * v[i] % mo;
  for (int j = 2; j <= (1 << n); j <<= 1)
    for (int i = 0; i < (1 << n); i += j)
      for (register int k = 0; k <= (j >> 1) - 1; k++) {
        int x = a[i + k], y = a[i + k + (j >> 1)];
        a[i + k] = 1LL * (x + y) * inv2 % mo;
        a[i + k + (j >> 1)] = 1LL * (x - y + mo) * inv2 % mo;
      }
  mi[0] = 1;
  for (register int i = 1; i <= n; i++) mi[i] = mi[i - 1] << 1;
  for (register int i = 1; i <= all; i++) cnt[i] = cnt[i >> 1] + (i & 1);
  for (register int i = 0; i <= all; i++)
    res = (res + 1LL * mi[n - cnt[i]] * a[i]) % mo;
  res = 3LL * res % mo;
  printf("%d", res);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
const int mo = 1e9 + 7, N = 1111111, inv2 = 5e8 + 4;
int n, all, v[N], a[N], cnt[N], mi[22], res;
char t[N];
int main() {
  scanf("%d", &n);
  all = (1 << n) - 1;
  scanf("%s", t);
  for (register int i = 0; i <= all; i++) v[i] = t[i] - '0';
  for (int j = 2; j <= (1 << n); j <<= 1)
    for (int i = 0; i < (1 << n); i += j)
      for (register int k = 0; k <= (j >> 1) - 1; k++) {
        int x = v[i + k], y = v[i + k + (j >> 1)];
        v[i + k] = x + y < mo ? x + y : x + y - mo;
        v[i + k + (j >> 1)] = x - y >= 0 ? x - y : x - y + mo;
      }
  for (register int i = 0; i <= all; i++) a[i] = 1LL * v[i] * v[i] % mo;
  for (int j = 2; j <= (1 << n); j <<= 1)
    for (int i = 0; i < (1 << n); i += j)
      for (register int k = 0; k <= (j >> 1) - 1; k++) {
        int x = a[i + k], y = a[i + k + (j >> 1)];
        a[i + k] = 1LL * (x + y) * inv2 % mo;
        a[i + k + (j >> 1)] = 1LL * (x - y + mo) * inv2 % mo;
      }
  mi[0] = 1;
  for (register int i = 1; i <= n; i++) mi[i] = mi[i - 1] << 1;
  for (register int i = 1; i <= all; i++) cnt[i] = cnt[i >> 1] + (i & 1);
  for (register int i = 0; i <= all; i++)
    res = (res + 1LL * mi[n - cnt[i]] * a[i]) % mo;
  res = 3LL * res % mo;
  printf("%d", res);
  return 0;
}

```