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 ll = long long;
int const modulo = 1000000007;
int const nmax = 20;
int lgpow(int a, int b) {
  if (b == 0)
    return 1;
  else if (b == 1)
    return a;
  else {
    int result = lgpow(a, b / 2);
    if (b % 2 == 0)
      return 1LL * result * result % modulo;
    else
      return 1LL * result * result % modulo * a % modulo;
  }
}
int dp[1 << nmax], dp2[1 << nmax];
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  int n;
  std::cin >> n;
  std::string s;
  std::cin >> s;
  for (int mask = 0; mask < (1 << n); mask++) {
    int val;
    val = s[mask] - '0';
    if (val == 0)
      dp[mask]++;
    else
      dp2[mask]++;
  }
  for (int i = 0; i < n; i++) {
    for (int mask = 0; mask < (1 << n); mask++) {
      if (0 < (mask & (1 << i))) {
        int mask2 = mask ^ (1 << i);
        int a = dp[mask], b = dp[mask2];
        dp[mask] = (a + 1LL * b * 2) % modulo;
        dp[mask2] = (b + 1LL * a * 2) % modulo;
      }
    }
  }
  int result = 0;
  for (int mask = 0; mask < (1 << n); mask++) {
    result += 1LL * dp[mask] * dp2[mask] % modulo;
    if (modulo <= result) result -= modulo;
  }
  std::cout << 1LL * result * 3 % modulo;
  return 0;
}
