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 MAXN = (1 << 20) + 10;
const long long MOD = 1e9 + 7;
inline long long pw(long long x, long long k) {
  long long res = 1;
  for (long long bs = x; k; k >>= 1, bs = (bs * bs) % MOD) {
    if (k & 1) res = (res * bs) % MOD;
  }
  return res;
}
inline long long Inv(long long x) { return pw(x, MOD - 2); }
inline void fwt(long long x[MAXN], int N, bool inv = 0) {
  for (int d = 1; d < N; d <<= 1) {
    int d2 = d << 1;
    for (int s = 0; s < N; s += d2) {
      for (int i = s, j = s + d; i < s + d; i++, j++) {
        long long ta = x[i], tb = x[j];
        x[i] = ta + tb;
        x[j] = ta - tb;
        if (x[i] >= MOD) x[i] -= MOD;
        if (x[j] < 0) x[j] += MOD;
      }
    }
  }
  if (inv)
    for (int i = 0; i < N; i++) {
      x[i] *= Inv(N);
      x[i] %= MOD;
    }
}
char buf[(1 << 20) + 20];
long long a[1 << (20 + 1)], m;
int main() {
  scanf("%lld", &m);
  scanf("%s", buf);
  for (int i = 0; i < (1 << m); i++) a[i] = buf[i] - '0';
  fwt(a, 1 << m);
  for (int i = 0; i < (1 << m); i++) a[i] = a[i] * a[i] % MOD;
  fwt(a, 1 << m, true);
  long long ans = 0;
  for (int i = 0; i < (1 << m); i++)
    ans = (ans + a[i] * (1 << (m - __builtin_popcount(i))) % MOD) % MOD;
  ans = ans * 3 % MOD;
  printf("%lld\n", ans);
}
