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>
char ch;
bool fs;
void re(int& x) {
  while (ch = getchar(), ch < 33)
    ;
  if (ch == '-')
    fs = 1, x = 0;
  else
    fs = 0, x = ch - 48;
  while (ch = getchar(), ch > 33) x = x * 10 + ch - 48;
  if (fs) x = -x;
}
using namespace std;
const int mod = 1000000007, inv2 = (mod + 1) / 2;
int n, m, ans, bit[1 << 20], a[1 << 20];
char s[1 << 20];
int add(int a) { return a < mod ? a : a - mod; }
int dec(int a) { return a < 0 ? a + mod : a; }
void fwt(int* x, int n, int f) {
  for (int i = 1; i < n; i <<= 1)
    for (int j = 0; j < n; j += i << 1)
      for (int k = 0; k < i; ++k) {
        int l = x[j + k], r = x[i + j + k];
        x[j + k] = add(l + r);
        x[i + j + k] = dec(l - r);
        if (!~f)
          x[j + k] = 1ll * x[j + k] * inv2 % mod,
                x[i + j + k] = 1ll * x[i + j + k] * inv2 % mod;
      }
}
int main() {
  re(n);
  m = 1 << n;
  scanf("%s", s);
  for (int i = 0; i < m; ++i) a[i] = s[i] - '0';
  fwt(a, m, 1);
  for (int i = 0; i < m; ++i) a[i] = 1ll * a[i] * a[i] % mod;
  fwt(a, m, -1);
  bit[0] = 1 << n;
  for (int i = 1; i < m; ++i) {
    bit[i] = bit[i >> 1];
    if (i & 1) bit[i] = 1ll * bit[i] * inv2 % mod;
  }
  for (int i = 0; i < m; ++i) ans = (ans + 1ll * a[i] * bit[i]) % mod;
  printf("%d\n", 3ll * ans % mod);
}
