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;
int n, g[1048576], pw[25], cnt[1048576];
void FWT(int *f, int type) {
  for (register int i = 1; i < n; i <<= 1)
    for (register int j = 0; j < n; j += i << 1)
      for (register int k = 0; k < i; ++k) {
        int x = f[j + k], y = f[i + j + k];
        f[j + k] = 1ll * type * (x + y) % 1000000007,
              f[i + j + k] = 1ll * type * (x + 1000000007 - y) % 1000000007;
      }
}
template <typename T>
inline void read(T &t) {
  t = 0;
  char c = getchar();
  int f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    t = (t << 3) + (t << 1) + c - '0';
    c = getchar();
  }
  t *= f;
}
template <typename T, typename... Args>
inline void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
signed main() {
  read(n);
  pw[n] = 1;
  for (register int i = n - 1; ~i; --i) pw[i] = pw[i + 1] * 2;
  n = 1 << n;
  for (register int i = 0; i < n; ++i)
    scanf("%1d", &g[i]), cnt[i] = cnt[i >> 1] + (i & 1);
  FWT(g, 1);
  for (register int i = 0; i < n; ++i) g[i] = 1ll * g[i] * g[i] % 1000000007;
  FWT(g, (1000000007 + 1) >> 1);
  int ans = 0;
  for (register int i = 0; i < n; ++i)
    ans = (ans + 1ll * g[i] * pw[cnt[i]] % 1000000007) % 1000000007;
  write(ans * 3ll % 1000000007), putchar('\n');
  return 0;
}
