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 mod = 1e9 + 7;
const int N = 1 << 20;
int a[N], b[N], c[N], d[N];
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % mod;
    }
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
struct FWT {
  int mod = 1e9 + 7;
  int rev = qpow(2, mod - 2);
  int add(int a, int b) {
    if (a + b >= mod) return a + b - mod;
    if (a + b < 0) return a + b + mod;
    return a + b;
  }
  int mul(int a, int b) { return (long long)a * b % mod; }
  void fwt(int a[], int n, int op) {
    for (int d = 1; d < n; d <<= 1) {
      for (int m = d << 1, i = 0; i < n; i += m) {
        for (int j = 0; j < d; j++) {
          int x = a[i + j], y = a[i + j + d];
          if (op == 0)
            a[i + j] = add(x, y), a[i + j + d] = add(x, -y);
          else
            a[i + j] = mul(add(x, y), rev), a[i + j + d] = mul(add(x, -y), rev);
        }
      }
    }
  }
  void solve(int a[], int b[], int n) {
    fwt(a, n, 0);
    fwt(b, n, 0);
    for (int i = (0); i < (n); i++) a[i] = mul(a[i], b[i]);
    fwt(a, n, 1);
  }
} Fwt;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int t = n;
  n = 1 << n;
  string s;
  cin >> s;
  for (int i = (0); i < (n); i++) {
    c[i] = s[i] ^ 48;
    d[i] = d[i >> 1] + (i & 1);
  }
  Fwt.fwt(c, n, 0);
  for (int i = (0); i < (n); i++) c[i] = (long long)c[i] * c[i] % mod;
  Fwt.fwt(c, n, 1);
  int ans = 0;
  for (int i = (0); i < (n); i++) {
    ans = (ans + (long long)c[i] * (1 << (t - d[i]))) % mod;
  }
  cout << (long long)ans * 3 % mod;
  return 0;
}
