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 = 2 * 1000 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int inv = 500 * 1000 * 1000 + 4;
void fwht(int n, int a[], bool inv) {
  if (n == 1) return;
  int m = n / 2, x, y;
  fwht(m, a, inv);
  fwht(m, a + m, inv);
  for (int i = 0; i < (m); ++i) {
    x = a[i];
    y = a[i + m];
    a[i] = (x + y) % mod;
    a[i + m] = (x - y + mod) % mod;
    if (inv)
      a[i] = (1LL * a[i] * ::inv) % mod,
      a[i + m] = (1LL * a[i + m] * ::inv) % mod;
  }
}
void mult(int n, int a[]) {
  fwht(n, a, false);
  for (int i = 0; i < (n); ++i) a[i] = 1LL * a[i] * a[i] % mod;
  fwht(n, a, true);
}
int n;
char a[N];
int b[N];
int result;
int m;
void sub() {
  for (int i = 0; i < (n); ++i) b[i] = a[i];
  mult(n, b);
  for (int i = 0; i < (n); ++i) {
    result = (result + b[i] * (3LL << (m - __builtin_popcount(i)))) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> a;
  m = n;
  n = 1 << n;
  for (int i = 0; i < (n); ++i) a[i] -= '0';
  sub();
  cout << result << '\n';
  return 0;
}
