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 N = 1 << 20 | 1;
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int a[N], s[N];
const int mod = 1e9 + 7;
inline void DFT(int n) {
  int i, o, k, x, y;
  for (i = 1; i < n; i <<= 1)
    for (o = 0; o < n; o += i << 1)
      for (k = 0; k < i; k++) {
        x = a[o + k], y = a[o + k + i];
        a[o + k] = (x + y) % mod;
        a[o + k + i] = (x - y) % mod;
      }
}
int main() {
  int i, ans = 0, n = 1 << gi();
  char c;
  const int inv2 = (mod + 1) >> 1;
  for (i = 0; i < n; a[i++] = c == '1')
    while ((c = getchar()) != '0' && c != '1')
      ;
  DFT(n);
  for (i = 0; i < n; i++) a[i] = 1LL * a[i] * a[i] % mod;
  DFT(n);
  s[0] = 1;
  for (i = 0; i < n; i++) {
    s[i] = s[i >> 1];
    if (i & 1) s[i] = 1LL * s[i] * inv2 % mod;
    ans = (ans + 1LL * s[i] * a[i]) % mod;
  }
  ans = 3LL * ans % mod;
  cout << (ans + mod) % mod << endl;
  return 0;
}
