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;
namespace FGF {
int n, m;
const int N = 2e6 + 5;
const int mo = 1e9 + 7;
char s[N];
long long a[N], pw[N], cnt[N];
long long ans;
void FWTxor(long long *y, int op) {
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < (1 << n); j += (1 << i))
      for (int k = 0; k < (1 << (i - 1)); k++) {
        long long c = (y[j | k] + y[j | k | (1 << (i - 1))]) % mo,
                  d = (y[j | k] - y[j | k | (1 << (i - 1))] + mo) % mo;
        y[j | k] = c * op % mo, y[j | k | (1 << (i - 1))] = d * op % mo;
      }
}
void work() {
  scanf("%d", &n);
  scanf("%s", s);
  pw[0] = 1;
  for (int i = 0; i < (1 << n); i++) a[i] = s[i] - '0';
  for (int i = 1; i <= n; i++) pw[i] = 2LL * pw[i - 1] % mo;
  for (int i = 1; i < (1 << n); i++) cnt[i] = cnt[i >> 1] + (i & 1);
  FWTxor(a, 1);
  for (int i = 0; i < (1 << n); i++) a[i] = a[i] * a[i] % mo;
  FWTxor(a, (mo + 1) / 2);
  for (int i = 0; i < (1 << n); i++)
    ans = (ans + 1LL * a[i] * pw[n - cnt[i]]) % mo;
  printf("%lld", ans * 3 % mo);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}
