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 = 1 << 20 | 1, mod = 1e9 + 7;
char st[N];
int n, ans, a[N];
inline void check(int& x) { x -= mod, x += x >> 31 & mod; }
inline int f(int x) { return 1 << __builtin_popcount(x); }
inline int Pow(long long int t, int x, long long int s = 1) {
  for (; x; x >>= 1, t = t * t % mod)
    if (x & 1) s = s * t % mod;
  return s;
}
inline void FWT(int* a, int l) {
  for (int i = 1, x, y; i < l; i <<= 1)
    for (int j = 0, d = i << 1; j < l; j += d)
      for (int k = 0; k <= i - 1; k++)
        x = a[j | k], y = a[i | j | k], check(a[j | k] = x + y),
        check(a[i | j | k] = x + mod - y);
}
inline void mul(int* a, int x, int l, int* c) {
  while (l--) *c++ = 1ll * x * *a++ % mod;
}
inline void dot(int* a, int* b, int l, int* c) {
  while (l--) *c++ = 1ll * *a++ * *b++ % mod;
}
inline void IFWT(int* a, int l) { FWT(a, l), mul(a, Pow(l, mod - 2), l, a); }
int main() {
  scanf("%d%s", &n, st), n = 1 << n;
  for (int i = 0; i <= n - 1; i++) a[i] = st[i] - '0';
  FWT(a, n), dot(a, a, n, a), IFWT(a, n);
  for (int i = 0; i <= n - 1; i++)
    check(ans += 1ll * f(i ^ n - 1) * a[i] % mod);
  return cout << 3ll * ans % mod, 0;
}
