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 inf = 1e9 + 7;
const int mod = inf;
const int maxn = 2e6 + 100;
long long inv2, n;
int cnt[maxn];
long long a[maxn];
long long qpow(long long x, long long a) {
  long long base = x, rt = 1;
  while (a) {
    if (a & 1) rt *= base, rt %= mod;
    base *= base, base %= mod;
    a >>= 1;
  }
  return rt;
}
void FWT_XOR(long long *a, int opt) {
  for (int i = 1; i < n; i <<= 1)
    for (int p = i << 1, j = 0; j < n; j += p)
      for (int k = 0; k < i; k++) {
        int X = a[j + k], Y = a[i + j + k];
        a[j + k] = (X + Y) % mod;
        a[i + j + k] = (X + mod - Y) % mod;
        if (opt == -1) {
          a[j + k] = 1ll * a[j + k] * inv2 % mod;
          a[i + j + k] = 1ll * a[i + j + k] * inv2 % mod;
        }
      }
}
void solve() {
  string s;
  cin >> n >> s;
  int N = n;
  n = 1 << n;
  for (int i = 0; i < n; i++) a[i] = s[i] - '0';
  for (int i = 1; i <= n; i++) cnt[i] = cnt[i >> 1] + (i & 1);
  FWT_XOR(a, 1);
  for (int i = 0; i <= n; i++) a[i] *= a[i], a[i] %= mod;
  FWT_XOR(a, -1);
  long long ans = 0;
  for (int i = 0; i <= n; i++) {
    ans += ((1 << (N - cnt[i])) * a[i]) % mod;
    ans %= mod;
  }
  ans *= 3, ans %= mod;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  int t = 1;
  inv2 = qpow(2, mod - 2);
  while (t--) {
    solve();
  }
}
