#include <bits/stdc++.h>
using namespace std;
const int N = (1 << 20) + 10;
const int mod = 1e9 + 7;
const int inv2 = (mod + 1) >> 1;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, len, f[N], g[N], bin[N];
char s[N];
void fwt(int *a) {
  for (int i = 1; i < len; i <<= 1)
    for (int p = i << 1, j = 0; j < len; j += p)
      for (int k = 0; k < i; k++) {
        int x = a[j + k], y = a[j + k + i];
        a[j + k] = (x + y) % mod, a[j + k + i] = (x - y + mod) % mod;
      }
}
void ifwt(int *a) {
  for (int i = 1; i < len; i <<= 1)
    for (int p = i << 1, j = 0; j < len; j += p)
      for (int k = 0; k < i; k++) {
        int x = a[j + k], y = a[j + k + i];
        a[j + k] = 1ll * (x + y) * inv2 % mod,
              a[j + k + i] = 1ll * (x - y + mod) * inv2 % mod;
      }
}
int main() {
  cin >> n;
  bin[0] = 1;
  for (int i = 1; i <= n; i++) bin[i] = (bin[i - 1] << 1) % mod;
  len = 1 << n;
  scanf("%s", s);
  for (int i = 0; i < len; i++) f[i] = g[i] = s[i] - '0';
  fwt(f), fwt(g);
  for (int i = 0; i < len; i++) f[i] = 1ll * f[i] * g[i] % mod;
  ifwt(f);
  int ans = 0;
  for (int i = 0; i < len; i++)
    ans = (ans + 1ll * f[i] * bin[n - __builtin_popcount(i)]) % mod;
  cout << 3ll * ans % mod;
  return 0;
}

### Prompt

Generate 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 = (1 << 20) + 10;
const int mod = 1e9 + 7;
const int inv2 = (mod + 1) >> 1;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, len, f[N], g[N], bin[N];
char s[N];
void fwt(int *a) {
  for (int i = 1; i < len; i <<= 1)
    for (int p = i << 1, j = 0; j < len; j += p)
      for (int k = 0; k < i; k++) {
        int x = a[j + k], y = a[j + k + i];
        a[j + k] = (x + y) % mod, a[j + k + i] = (x - y + mod) % mod;
      }
}
void ifwt(int *a) {
  for (int i = 1; i < len; i <<= 1)
    for (int p = i << 1, j = 0; j < len; j += p)
      for (int k = 0; k < i; k++) {
        int x = a[j + k], y = a[j + k + i];
        a[j + k] = 1ll * (x + y) * inv2 % mod,
              a[j + k + i] = 1ll * (x - y + mod) * inv2 % mod;
      }
}
int main() {
  cin >> n;
  bin[0] = 1;
  for (int i = 1; i <= n; i++) bin[i] = (bin[i - 1] << 1) % mod;
  len = 1 << n;
  scanf("%s", s);
  for (int i = 0; i < len; i++) f[i] = g[i] = s[i] - '0';
  fwt(f), fwt(g);
  for (int i = 0; i < len; i++) f[i] = 1ll * f[i] * g[i] % mod;
  ifwt(f);
  int ans = 0;
  for (int i = 0; i < len; i++)
    ans = (ans + 1ll * f[i] * bin[n - __builtin_popcount(i)]) % mod;
  cout << 3ll * ans % mod;
  return 0;
}

```