#include <bits/stdc++.h>
using namespace std;
const int MAXN = (1 << 20) + 5;
const int Mod = 1000000007;
int fp(int a, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1LL * res * a % Mod;
    a = 1LL * a * a % Mod;
    k >>= 1;
  }
  return res;
}
void fwt(int f[], int n, int on) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < (1 << n); j++) {
      if (j & (1 << i)) continue;
      int u = f[j], v = f[j | (1 << i)];
      f[j] = (u + v) % Mod, f[j | (1 << i)] = (u - v + Mod) % Mod;
    }
  if (on == -1) {
    int t = fp(1 << n, Mod - 2);
    for (int i = 0; i < (1 << n); i++) f[i] = 1LL * t * f[i] % Mod;
  }
}
char str[MAXN];
int f[MAXN], c[MAXN];
int main() {
  int n;
  scanf("%d%s", &n, str);
  for (int i = 0; i < (1 << n); i++) {
    f[i] = (str[i] - '0');
    c[i] = n - __builtin_popcount(i);
  }
  fwt(f, n, 1);
  for (int i = 0; i < (1 << n); i++) f[i] = 1LL * f[i] * f[i] % Mod;
  fwt(f, n, -1);
  int res = 0;
  for (int i = 0; i < (1 << n); i++)
    res = (res + 1LL * f[i] * (1 << c[i]) % Mod) % Mod;
  res = 3LL * res % Mod;
  printf("%d\n", res);
  return 0;
}

### Prompt

Please create a solution in CPP 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 MAXN = (1 << 20) + 5;
const int Mod = 1000000007;
int fp(int a, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1LL * res * a % Mod;
    a = 1LL * a * a % Mod;
    k >>= 1;
  }
  return res;
}
void fwt(int f[], int n, int on) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < (1 << n); j++) {
      if (j & (1 << i)) continue;
      int u = f[j], v = f[j | (1 << i)];
      f[j] = (u + v) % Mod, f[j | (1 << i)] = (u - v + Mod) % Mod;
    }
  if (on == -1) {
    int t = fp(1 << n, Mod - 2);
    for (int i = 0; i < (1 << n); i++) f[i] = 1LL * t * f[i] % Mod;
  }
}
char str[MAXN];
int f[MAXN], c[MAXN];
int main() {
  int n;
  scanf("%d%s", &n, str);
  for (int i = 0; i < (1 << n); i++) {
    f[i] = (str[i] - '0');
    c[i] = n - __builtin_popcount(i);
  }
  fwt(f, n, 1);
  for (int i = 0; i < (1 << n); i++) f[i] = 1LL * f[i] * f[i] % Mod;
  fwt(f, n, -1);
  int res = 0;
  for (int i = 0; i < (1 << n); i++)
    res = (res + 1LL * f[i] * (1 << c[i]) % Mod) % Mod;
  res = 3LL * res % Mod;
  printf("%d\n", res);
  return 0;
}

```