#include <bits/stdc++.h>
using namespace std;
const int maxN = 20;
const int maxM = (1 << maxN) + 10;
const int Mod = 1000000007;
const int inv2 = 500000004;
int n;
char In[maxM];
int A[maxM], C[maxM], pw[maxN + 10];
void FWT(int N, int *P, int opt);
int main() {
  scanf("%d", &n);
  int N = 1 << n;
  scanf("%s", In);
  for (int i = 0; i < N; i++) A[i] = In[i] - '0';
  FWT(N, A, 1);
  for (int i = 0; i < N; i++) A[i] = 1ll * A[i] * A[i] % Mod;
  FWT(N, A, -1);
  pw[0] = 1;
  for (int i = 1; i < maxN + 10; i++) pw[i] = 2ll * pw[i - 1] % Mod;
  for (int i = 0; i < N; i++) C[i] = C[i >> 1] + (i & 1);
  int Ans = 0;
  for (int i = 0; i < N; i++)
    Ans = (Ans + 1ll * pw[n - C[i]] * A[i] % Mod) % Mod;
  Ans = 3ll * Ans % Mod;
  printf("%d\n", Ans);
  return 0;
}
void FWT(int N, int *P, int opt) {
  for (int i = 1; i < N; i <<= 1)
    for (int j = 0; j < N; j += (i << 1))
      for (int k = 0; k < i; k++) {
        int X = P[j + k], Y = P[j + k + i];
        P[j + k] = (X + Y) % Mod;
        P[j + k + i] = (X - Y + Mod) % Mod;
        if (opt == -1)
          P[j + k] = 1ll * P[j + k] * inv2 % Mod,
                P[j + k + i] = 1ll * P[j + k + i] * inv2 % Mod;
      }
  return;
}

### Prompt

Please formulate 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 maxN = 20;
const int maxM = (1 << maxN) + 10;
const int Mod = 1000000007;
const int inv2 = 500000004;
int n;
char In[maxM];
int A[maxM], C[maxM], pw[maxN + 10];
void FWT(int N, int *P, int opt);
int main() {
  scanf("%d", &n);
  int N = 1 << n;
  scanf("%s", In);
  for (int i = 0; i < N; i++) A[i] = In[i] - '0';
  FWT(N, A, 1);
  for (int i = 0; i < N; i++) A[i] = 1ll * A[i] * A[i] % Mod;
  FWT(N, A, -1);
  pw[0] = 1;
  for (int i = 1; i < maxN + 10; i++) pw[i] = 2ll * pw[i - 1] % Mod;
  for (int i = 0; i < N; i++) C[i] = C[i >> 1] + (i & 1);
  int Ans = 0;
  for (int i = 0; i < N; i++)
    Ans = (Ans + 1ll * pw[n - C[i]] * A[i] % Mod) % Mod;
  Ans = 3ll * Ans % Mod;
  printf("%d\n", Ans);
  return 0;
}
void FWT(int N, int *P, int opt) {
  for (int i = 1; i < N; i <<= 1)
    for (int j = 0; j < N; j += (i << 1))
      for (int k = 0; k < i; k++) {
        int X = P[j + k], Y = P[j + k + i];
        P[j + k] = (X + Y) % Mod;
        P[j + k + i] = (X - Y + Mod) % Mod;
        if (opt == -1)
          P[j + k] = 1ll * P[j + k] * inv2 % Mod,
                P[j + k + i] = 1ll * P[j + k + i] * inv2 % Mod;
      }
  return;
}

```