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;
long long gi() {
  char cc = getchar();
  long long cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const long long P = 1e9 + 7;
const long long Iv2 = 5e8 + 4;
const long long N = (1 << 21);
long long n, limit, f[N], g[N], fac[N];
char s[N];
long long lowbit(long long x) { return __builtin_popcountll(x); }
void FWT_xor(long long *a, long long type) {
  for (register long long k = 1; k < limit; k <<= 1)
    for (register long long i = 0; i < limit; i = (i + (k << 1)))
      for (register long long j = i; j < i + k; ++j) {
        long long nx = a[j], ny = a[j + k];
        (type == 1)
            ? (a[j] = (nx + ny) % P, a[j + k] = (nx - ny + P) % P)
            : (a[j] = (nx + ny) * Iv2 % P, a[j + k] = (nx - ny + P) * Iv2 % P);
      }
}
signed main() {
  n = gi(), limit = (1 << n);
  scanf("%s", s);
  long long S = limit - 1;
  for (register long long i = (0); i <= (limit - 1); ++i) f[i] = s[i] - '0';
  for (long long i = 0; i < limit; ++i) g[i] = f[i ^ S];
  FWT_xor(f, 1), FWT_xor(g, 1);
  for (register long long i = (0); i <= (limit); ++i) g[i] = f[i] * g[i] % P;
  FWT_xor(g, 0);
  long long Ans = 0;
  fac[0] = 1;
  for (register long long i = (1); i <= (25); ++i) fac[i] = fac[i - 1] * 2 % P;
  for (register long long i = (0); i <= (limit); ++i)
    Ans = (Ans + g[i] * fac[lowbit(i)] % P) % P;
  cout << (Ans * 3) % P << endl;
  return 0;
}
