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;
template <typename T>
inline bool chkmin(T &x, T y) {
  return (y < x) ? (x = y, 1) : 0;
}
template <typename T>
inline bool chkmax(T &x, T y) {
  return (y > x) ? (x = y, 1) : 0;
}
inline int read() {
  int x;
  char c;
  int f = 1;
  while ((c = getchar()) != '-' && (c > '9' || c < '0'))
    ;
  if (c == '-') f = -1, c = getchar();
  x = c ^ '0';
  while ((c = getchar()) >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ '0');
  return x * f;
}
inline long long readll() {
  long long x;
  char c;
  int f = 1;
  while ((c = getchar()) != '-' && (c > '9' || c < '0'))
    ;
  if (c == '-') f = -1, c = getchar();
  x = c ^ '0';
  while ((c = getchar()) >= '0' && c <= '9')
    x = (x << 1ll) + (x << 3ll) + (c ^ '0');
  return x * f;
}
const int mod = 1e9 + 7;
char s[1 << 20];
int Pw[30], n, a[1 << 20], N, inv2 = (mod + 1) / 2;
void fwt_xor(int *y, int flag) {
  for (int k = 2; k <= N; k <<= 1)
    for (int i = 0; i < N; i += k)
      for (int j = 0; j < (k >> 1); ++j) {
        int x1 = y[i + j], x2 = y[i + j + (k >> 1)];
        if (flag)
          y[i + j] = (x1 + x2) % mod,
                y[i + j + (k >> 1)] = (x1 - x2 + mod) % mod;
        else
          y[i + j] = (long long)(x1 + x2) * inv2 % mod,
                y[i + j + (k >> 1)] = (long long)(x1 - x2 + mod) * inv2 % mod;
      }
}
int main() {
  n = read();
  scanf("%s", s);
  N = (1 << n);
  for (register int i = 0, iend = N - 1; i <= iend; ++i) a[i] = s[i] - '0';
  fwt_xor(a, 1);
  for (register int i = 0, iend = N - 1; i <= iend; ++i)
    a[i] = (long long)a[i] * a[i] % mod;
  fwt_xor(a, 0);
  Pw[0] = 1;
  for (register int i = 1, iend = n; i <= iend; ++i)
    Pw[i] = (Pw[i - 1] << 1) % mod;
  int ans = 0;
  for (register int i = 0, iend = N - 1; i <= iend; ++i)
    ans = (ans + (long long)Pw[n - __builtin_popcount(i)] * a[i]) % mod;
  ans = (long long)ans * 3 % mod;
  printf("%d\n", ans);
  return 0;
}
