#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
char s[1 << 20];
int a[1 << 20];
int qp[25];
int b[1 << 20];
int n;
int quick_pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    b >>= 1;
    a = 1ll * a * a % mod;
  }
  return res;
}
void fwt_xor(int* a, int l, int opt) {
  for (int i = 1; i < l; i <<= 1) {
    for (int j = 0; j < l; j += (i << 1)) {
      for (int k = 0; k < i; k++) {
        int tmp = a[j + k];
        a[j + k] = (tmp + a[j + k + i]) % mod;
        a[j + k + i] = (tmp + mod - a[j + k + i]) % mod;
      }
    }
  }
  if (opt == -1) {
    int inv = quick_pow(l, mod - 2);
    for (int i = 0; i < l; i++) a[i] = 1ll * a[i] * inv % mod;
  }
}
int main() {
  scanf("%d", &n);
  scanf(" %s", s);
  qp[0] = 1;
  for (int i = 1; i <= n; i++) qp[i] = qp[i - 1] * 2;
  for (int i = 1; i < (1 << n); i++) b[i] = b[i >> 1] + (i & 1);
  for (int i = 0; i < (1 << n); i++) a[i] = s[i] - '0';
  fwt_xor(a, 1 << n, 1);
  for (int i = 0; i < (1 << n); i++) a[i] = 1ll * a[i] * a[i] % mod;
  fwt_xor(a, 1 << n, -1);
  int ans = 0;
  for (int i = 0; i < (1 << n); i++)
    ans += 1ll * a[i] * qp[n - b[i]] % mod, ans %= mod;
  ans = 1ll * ans * 3 % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 mod = 1e9 + 7;
char s[1 << 20];
int a[1 << 20];
int qp[25];
int b[1 << 20];
int n;
int quick_pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    b >>= 1;
    a = 1ll * a * a % mod;
  }
  return res;
}
void fwt_xor(int* a, int l, int opt) {
  for (int i = 1; i < l; i <<= 1) {
    for (int j = 0; j < l; j += (i << 1)) {
      for (int k = 0; k < i; k++) {
        int tmp = a[j + k];
        a[j + k] = (tmp + a[j + k + i]) % mod;
        a[j + k + i] = (tmp + mod - a[j + k + i]) % mod;
      }
    }
  }
  if (opt == -1) {
    int inv = quick_pow(l, mod - 2);
    for (int i = 0; i < l; i++) a[i] = 1ll * a[i] * inv % mod;
  }
}
int main() {
  scanf("%d", &n);
  scanf(" %s", s);
  qp[0] = 1;
  for (int i = 1; i <= n; i++) qp[i] = qp[i - 1] * 2;
  for (int i = 1; i < (1 << n); i++) b[i] = b[i >> 1] + (i & 1);
  for (int i = 0; i < (1 << n); i++) a[i] = s[i] - '0';
  fwt_xor(a, 1 << n, 1);
  for (int i = 0; i < (1 << n); i++) a[i] = 1ll * a[i] * a[i] % mod;
  fwt_xor(a, 1 << n, -1);
  int ans = 0;
  for (int i = 0; i < (1 << n); i++)
    ans += 1ll * a[i] * qp[n - b[i]] % mod, ans %= mod;
  ans = 1ll * ans * 3 % mod;
  printf("%d\n", ans);
  return 0;
}

```