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>
using namespace std;
inline int read() {
  char x = 0;
  while (x < '0' || x > '9') x = getchar();
  return x - '0';
}
const int mod = 1e9 + 7;
inline int add(int a, int b) {
  a += b;
  return a >= mod ? a - mod : a;
}
inline int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + mod : a;
}
inline int mul(int a, int b) { return (long long)a * b % mod; }
inline int qpow(int a, int b) {
  int ret = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) ret = mul(ret, a);
  return ret;
}
const int inv2 = qpow(2, mod - 2);
const int N = 1.5e6 + 5;
int A[N], B[N], n, len;
inline void FWT_xor(int *t, int n, int type) {
  for (int step = 1; step < n; step <<= 1)
    for (int i = 0; i < n; i += step << 1)
      for (int j = 0; j < step; j++) {
        int x = t[i + j], y = t[i + j + step];
        t[i + j] = add(x, y), t[i + j + step] = sub(x, y);
        if (type == -1)
          t[i + j] = mul(t[i + j], inv2),
                t[i + j + step] = mul(t[i + j + step], inv2);
      }
}
int main() {
  scanf("%d", &n);
  len = 1 << n;
  for (int i = 0; i < 1 << n; i++) {
    A[i] = read();
    B[i] = A[i] ^ 1;
  }
  FWT_xor(A, len, 1);
  FWT_xor(B, len, 1);
  for (int i = 0; i < len; i++) A[i] = mul(A[i], B[i]);
  FWT_xor(A, len, -1);
  int ans = 0;
  for (int i = 0; i < len; i++) {
    int Q = __builtin_popcount(i);
    ans = add(ans, mul(1 << Q, A[i]));
  }
  printf("%d\n", mul(ans, 3));
}
