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 M = (1 << 20) + 2, mod = 1e9 + 7, inv2 = (mod + 1) / 2;
int n;
int f[M], ans;
char s[M];
void Add(int &x, int y) { x = (x + y) % mod; }
int add(int x, int y) { return (x + y) % mod; }
int sub(int x, int y) { return (x - y + mod) % mod; }
void Mul(int &x, int y) { x = 1ll * x * y % mod; }
int mul(int x, int y) { return 1ll * x * y % mod; }
int qpow(int a, int b) {
  int ret = 1;
  for (; b; b >>= 1, Mul(a, a))
    if (b & 1) Mul(ret, a);
  return ret;
}
void FWT(int a[], int op) {
  for (int i = 1; i < (1 << n); i <<= 1)
    for (int j = 0; j <= (1 << n) - 1; j++)
      if (!(j & i)) {
        int x = a[j], y = a[j + i];
        a[j] = add(x, y), a[j + i] = sub(x, y);
        if (op == -1) Mul(a[j], inv2), Mul(a[j + i], inv2);
      }
}
int main() {
  cin >> n;
  scanf("%s", s);
  for (int i = 0; i <= (1 << n) - 1; i++) f[i] = s[i] - '0';
  FWT(f, 1);
  for (int i = 0; i <= (1 << n) - 1; i++) Mul(f[i], f[i]);
  FWT(f, -1);
  for (int s = 0; s <= (1 << n) - 1; s++)
    Add(ans, mul(f[s], qpow(2, n - __builtin_popcount(s))));
  ans = (ans % mod + mod) % mod;
  printf("%d\n", mul(ans, 3));
  return 0;
}
