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;
const int e = (1 << 20) | 5, mod = 1e9 + 7;
int n, all, a[e], tot, cnt[e], ans;
char s[e];
inline void add(int &x, int y) { (x += y) >= mod && (x -= mod); }
inline void del(int &x, int y) { (x -= y) < 0 && (x += mod); }
inline int plu(int x, int y) {
  add(x, y);
  return x;
}
inline int sub(int x, int y) {
  del(x, y);
  return x;
}
inline int ksm(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = (long long)res * x % mod;
    y >>= 1;
    x = (long long)x * x % mod;
  }
  return res;
}
inline void fwt(int *a, int op) {
  int i, j, k;
  for (k = 1; k < all; k <<= 1)
    for (i = 0; i < all; i += k << 1)
      for (j = 0; j < k; j++) {
        int b = a[i + j], c = a[i + j + k];
        a[i + j] = plu(b, c);
        a[i + j + k] = sub(b, c);
      }
  if (op == -1) {
    for (i = 0; i < all; i++) a[i] = (long long)a[i] * tot % mod;
  }
}
int main() {
  cin >> n;
  int i;
  all = 1 << n;
  tot = ksm(all, mod - 2);
  scanf("%s", s);
  for (i = 0; i < all; i++) {
    cnt[i] = cnt[i >> 1] + (i & 1);
    if (s[i] == '1') a[i] = 1;
  }
  fwt(a, 1);
  for (i = 0; i < all; i++) a[i] = (long long)a[i] * a[i] % mod;
  fwt(a, -1);
  for (i = 0; i < all; i++)
    ans = (ans + (long long)a[i] * (1 << n - cnt[i])) % mod;
  ans = 3ll * ans % mod;
  cout << ans << endl;
  return 0;
}
