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;
inline int Read() {
  int x = 0, f = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 1048580;
const int mod = 1e9 + 7;
int n, m, ans, a[MAXN], b[MAXN];
char s[MAXN];
inline void FWT(int *x) {
  for (int i = 1; i < m; i <<= 1)
    for (int j = 0; j < m; j += i << 1)
      for (int k = 0, u, v; k < i; k++)
        u = x[j + k], v = x[j + k + i], x[j + k] = (u + v) % mod,
        x[j + k + i] = (u - v) % mod;
}
inline int Qow(int x, int y) {
  int r = 1;
  for (; y; y >>= 1, x = 1LL * x * x % mod)
    if (y & 1) r = 1LL * r * x % mod;
  return r;
}
int main() {
  n = Read(), m = 1 << n;
  scanf("%s", s);
  for (int i = 0; i < m; i++) a[i] = s[i] - '0', b[i] = a[i] ^ 1;
  FWT(a);
  FWT(b);
  for (int i = 0; i < m; i++) a[i] = 1LL * a[i] * b[i] % mod;
  FWT(a);
  for (int i = 0, v = Qow(m, mod - 2); i < m; i++) a[i] = 1LL * a[i] * v % mod;
  for (int i = 0; i < m; i++)
    ans = (ans + (1LL << __builtin_popcount(i)) * a[i]) % mod;
  return printf("%d\n", (3LL * ans) % mod), 0;
}
