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.
#include <bits/stdc++.h>
const int MOD = 1000000007;
const int Inv2 = 500000004;
inline int mul(int x, int y) { return (long long)x * (long long)y % MOD; }
inline int add(int x, int y) {
  int r = x + y;
  if (r >= MOD) r -= MOD;
  return r;
}
inline int qpow(int a, int p) {
  int x = a, ans = 1;
  while (p) {
    if (p & 1) ans = mul(ans, x);
    x = mul(x, x);
    p >>= 1;
  }
  return ans;
}
using namespace std;
template <typename T>
inline void read(T &x) {
  char ch;
  while ((ch = getchar()), (ch < '0' || ch > '9'))
    ;
  x = ch - '0';
  while ((ch = getchar()), (ch >= '0' && ch <= '9')) x = x * 10 + (ch - '0');
}
inline int readbin(int *str) {
  char ch;
  int len = 1;
  while ((ch = getchar()), (ch < '0' || ch > '1'))
    ;
  *(str++) = ch - '0';
  while ((ch = getchar()), (ch >= '0' && ch <= '1')) *(str++) = ch - '0', len++;
  return len;
}
void fwt(int A[], int n) {
  int i, h, k, A0, A1;
  for (h = 2; h <= n; h <<= 1) {
    for (i = 0; i < n; i += h) {
      for (k = i; k < i + (h / 2); k++) {
        A0 = A[k];
        A1 = A[k + (h / 2)];
        A[k] = (A0 + A1) % MOD;
        A[k + (h / 2)] = (A0 - A1) % MOD;
      }
    }
  }
}
void ifwt(int A[], int n) {
  int i, h, k, A0, A1;
  for (h = 2; h <= n; h <<= 1) {
    for (i = 0; i < n; i += h) {
      for (k = i; k < i + (h / 2); k++) {
        A0 = A[k];
        A1 = A[k + (h / 2)];
        A[k] = mul(Inv2, A0 + A1);
        A[k + (h / 2)] = mul(Inv2, A0 - A1);
      }
    }
  }
}
const int MAXN = 5 + (1 << 20);
int A[MAXN], P[MAXN];
int N, NB;
int main() {
  int i;
  read(NB);
  N = readbin(A);
  fwt(A, N);
  for (i = 0; i < N; i++) P[i] = mul(A[i], A[i]);
  ifwt(P, N);
  int Ans = 0;
  for (i = 0; i < N; i++)
    Ans = add(Ans, mul(P[i], 1 << (NB - __builtin_popcount(i))));
  Ans = mul(Ans, 3);
  printf("%d", Ans);
}
