#include <bits/stdc++.h>
const int mod = 1e9 + 7;
using int64 = long long;
int inv(int x) {
  return x == 1 ? x : int64(mod - mod / x) * inv(mod % x) % mod;
}
void FWT(int a[], int n, bool inv = false) {
  for (int k = 0; (1 << k) < n; ++k) {
    for (int i = 0; i < n; ++i) {
      if (i >> k & 1) continue;
      int j = i ^ (1 << k);
      int y = (a[i] + a[j]) % mod;
      int x = (a[i] - a[j] + mod) % mod;
      if (inv) {
        x = (mod - x) % mod;
        std::swap(x, y);
      }
      a[i] = x, a[j] = y;
    }
  }
}
const int N = 1 << 20;
int a[N], b[N];
int main() {
  int n;
  std::string s;
  std::cin >> n >> s;
  for (int i = 0; i < (1 << n); ++i) {
    a[i] = s[i] - '0';
    b[i] = 1 - a[i];
  }
  FWT(a, 1 << n);
  FWT(b, 1 << n);
  for (int i = 0; i < (1 << n); ++i) {
    a[i] = 1ll * a[i] * b[i] % mod;
  }
  FWT(a, 1 << n, true);
  for (int i = 0; i < (1 << n); ++i) {
    a[i] = 1ll * a[i] * inv(1 << n) % mod;
  }
  int ret = 0;
  for (int i = 0; i < (1 << n); ++i) {
    ret = (ret + 1ll * a[i] * (1 << __builtin_popcount(i))) % mod;
  }
  std::cout << ret * 3ll % mod << std::endl;
  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>
const int mod = 1e9 + 7;
using int64 = long long;
int inv(int x) {
  return x == 1 ? x : int64(mod - mod / x) * inv(mod % x) % mod;
}
void FWT(int a[], int n, bool inv = false) {
  for (int k = 0; (1 << k) < n; ++k) {
    for (int i = 0; i < n; ++i) {
      if (i >> k & 1) continue;
      int j = i ^ (1 << k);
      int y = (a[i] + a[j]) % mod;
      int x = (a[i] - a[j] + mod) % mod;
      if (inv) {
        x = (mod - x) % mod;
        std::swap(x, y);
      }
      a[i] = x, a[j] = y;
    }
  }
}
const int N = 1 << 20;
int a[N], b[N];
int main() {
  int n;
  std::string s;
  std::cin >> n >> s;
  for (int i = 0; i < (1 << n); ++i) {
    a[i] = s[i] - '0';
    b[i] = 1 - a[i];
  }
  FWT(a, 1 << n);
  FWT(b, 1 << n);
  for (int i = 0; i < (1 << n); ++i) {
    a[i] = 1ll * a[i] * b[i] % mod;
  }
  FWT(a, 1 << n, true);
  for (int i = 0; i < (1 << n); ++i) {
    a[i] = 1ll * a[i] * inv(1 << n) % mod;
  }
  int ret = 0;
  for (int i = 0; i < (1 << n); ++i) {
    ret = (ret + 1ll * a[i] * (1 << __builtin_popcount(i))) % mod;
  }
  std::cout << ret * 3ll % mod << std::endl;
  return 0;
}

```