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>
const int oo = 2139063143;
const int N = 2010000;
const int P = 1000000007, inv2 = (P + 1) >> 1;
using namespace std;
inline void sc(int &x) {
  x = 0;
  static int p;
  p = 1;
  static char c;
  c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - 48);
    c = getchar();
  }
  x *= p;
}
inline void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
inline void pr(int x) { print(x), putchar('\n'); }
int G(int x) { return x >= P ? x - P : (x < 0 ? x + P : x); }
void fwt(int *a, int n, int kind) {
  for (int i = 1; i < n; i <<= 1)
    for (int j = 0, sb = i << 1; j < n; j += sb)
      for (int k = j; k < j + i; k++) {
        int x = a[k], y = a[k + i];
        a[k] = G(x + y), a[k + i] = G(x - y);
        if (kind == -1)
          a[k] = 1LL * a[k] * inv2 % P, a[k + i] = 1LL * a[k + i] * inv2 % P;
      }
}
int a[N], mi[N], cnt[N];
char sb[N];
int main() {
  int n, s;
  sc(n);
  s = 1 << n;
  scanf("%s", sb + 1);
  mi[0] = 1;
  for (int i = 1; i <= s; i++)
    a[i - 1] = sb[i] - '0', mi[i] = G(mi[i - 1] + mi[i - 1]),
          cnt[i] = cnt[i >> 1] + (i & 1);
  fwt(a, s, 1);
  for (int i = 0; i < s; i++) a[i] = 1LL * a[i] * a[i] % P;
  fwt(a, s, -1);
  int ans = 0;
  for (int i = 0; i < s; i++) ans = G(ans + 1LL * a[i] * mi[n - cnt[i]] % P);
  pr(3LL * ans % P);
  return 0;
}
