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) + 500;
const int MOD = 1e9 + 7;
int pod_2 = 1;
inline int add(int A, int B) {
  if (A + B >= MOD) return A + B - MOD;
  return A + B;
}
inline int sub(int A, int B) {
  if (A - B < 0) return A - B + MOD;
  return A - B;
}
inline int mul(int A, int B) { return (long long)A * B % MOD; }
int a[N], n;
void FWT(int l, int r) {
  if (l == r) return;
  int mid = (l + r) / 2;
  FWT(l, mid);
  FWT(mid + 1, r);
  for (int i = l; i <= mid; i++) {
    int x = a[i], y = a[i + (mid - l + 1)];
    a[i] = mul(add(x, y), pod_2);
    a[i + (mid - l + 1)] = mul(sub(x, y), pod_2);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (1 << n); i++) {
    char c;
    scanf(" %c", &c);
    a[i] = c - '0';
  }
  FWT(0, (1 << n) - 1);
  pod_2 = (MOD + 1) / 2;
  for (int i = 0; i < (1 << n); i++) a[i] = mul(a[i], a[i]);
  FWT(0, (1 << n) - 1);
  int sol = 0;
  for (int i = 0; i < (1 << n); i++) {
    int kol = __builtin_popcount(i);
    sol = add(sol, mul((1 << (n - kol)), a[i]));
  }
  printf("%d\n", mul(sol, 3));
  return 0;
}
