#include <bits/stdc++.h>
using namespace std;
const int N = 1100000, md = 1000000007;
int n, i, f[N], num[N], ans, pw2[N], inv2 = (md + 1) / 2;
char s[N];
void fwtxor(int a[N], int t) {
  for (int l = 2; l <= (1 << n); l <<= 1) {
    int m = l >> 1;
    for (int i = 0; i < (1 << n); i += l)
      for (int j = 0; j < m; ++j) {
        int x = a[i + j], y = a[i + j + m];
        a[i + j] = (x + y) % md;
        a[i + j + m] = (x - y + md) % md;
        if (t == -1)
          a[i + j] = 1ll * a[i + j] * inv2 % md,
                a[i + j + m] = 1ll * a[i + j + m] * inv2 % md;
      }
  }
}
int main() {
  scanf("%d%s", &n, s);
  pw2[0] = 1;
  for (i = 1; i <= n; ++i) pw2[i] = pw2[i - 1] * 2 % md;
  for (i = 1; i < (1 << n); ++i) num[i] = num[i ^ (i & -i)] + 1;
  for (i = 0; i < (1 << n); ++i)
    if (s[i] == '1') f[i] = 1;
  fwtxor(f, 1);
  for (i = 0; i < (1 << n); ++i) f[i] = 1ll * f[i] * f[i] % md;
  fwtxor(f, -1);
  for (i = 0; i < (1 << n); ++i)
    ans = (ans + 1ll * f[i] * pw2[n - num[i]]) % md;
  for (i = 0; i < (1 << n); ++i)
    if (s[i] == '0')
      f[i] = 1;
    else
      f[i] = 0;
  fwtxor(f, 1);
  for (i = 0; i < (1 << n); ++i) f[i] = 1ll * f[i] * f[i] % md;
  fwtxor(f, -1);
  for (i = 0; i < (1 << n); ++i)
    ans = (ans + 1ll * f[i] * pw2[n - num[i]] % md * 2) % md;
  printf("%d", ans);
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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>
using namespace std;
const int N = 1100000, md = 1000000007;
int n, i, f[N], num[N], ans, pw2[N], inv2 = (md + 1) / 2;
char s[N];
void fwtxor(int a[N], int t) {
  for (int l = 2; l <= (1 << n); l <<= 1) {
    int m = l >> 1;
    for (int i = 0; i < (1 << n); i += l)
      for (int j = 0; j < m; ++j) {
        int x = a[i + j], y = a[i + j + m];
        a[i + j] = (x + y) % md;
        a[i + j + m] = (x - y + md) % md;
        if (t == -1)
          a[i + j] = 1ll * a[i + j] * inv2 % md,
                a[i + j + m] = 1ll * a[i + j + m] * inv2 % md;
      }
  }
}
int main() {
  scanf("%d%s", &n, s);
  pw2[0] = 1;
  for (i = 1; i <= n; ++i) pw2[i] = pw2[i - 1] * 2 % md;
  for (i = 1; i < (1 << n); ++i) num[i] = num[i ^ (i & -i)] + 1;
  for (i = 0; i < (1 << n); ++i)
    if (s[i] == '1') f[i] = 1;
  fwtxor(f, 1);
  for (i = 0; i < (1 << n); ++i) f[i] = 1ll * f[i] * f[i] % md;
  fwtxor(f, -1);
  for (i = 0; i < (1 << n); ++i)
    ans = (ans + 1ll * f[i] * pw2[n - num[i]]) % md;
  for (i = 0; i < (1 << n); ++i)
    if (s[i] == '0')
      f[i] = 1;
    else
      f[i] = 0;
  fwtxor(f, 1);
  for (i = 0; i < (1 << n); ++i) f[i] = 1ll * f[i] * f[i] % md;
  fwtxor(f, -1);
  for (i = 0; i < (1 << n); ++i)
    ans = (ans + 1ll * f[i] * pw2[n - num[i]] % md * 2) % md;
  printf("%d", ans);
  return 0;
}

```