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;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(int a) {
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(int a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  write(a);
  puts("");
}
const int mod = 1000000007, rev = (mod + 1) / 2, N = 20;
int ans, n;
long long a[1 << N], ycl[21];
char ch[(1 << N) + 5];
void FWT(long long *a, int n) {
  for (int k = 1; k < n; k <<= 1) {
    for (int i = 0; i < n; i += k << 1) {
      for (int j = 0; j < k; j++) {
        long long x = a[i + j];
        a[i + j] = a[i + j] + a[i + j + k];
        a[i + j + k] = x - a[i + j + k];
      }
    }
  }
}
void UFWT(long long *a, int n) {
  for (int k = 1; k < n; k <<= 1) {
    for (int i = 0; i < n; i += k << 1) {
      for (int j = 0; j < k; j++) {
        long long x = a[i + j];
        a[i + j] = (a[i + j] + a[i + j + k]) % mod;
        a[i + j + k] = (x - a[i + j + k]) % mod;
      }
    }
  }
}
int main() {
  n = read();
  scanf("%s", ch);
  for (int i = 0; i < (1 << n); i++) a[i] = ch[i] - '0';
  FWT(a, 1 << n);
  for (int i = 0; i < (1 << n); i++) a[i] = a[i] * a[i] % mod;
  UFWT(a, 1 << n);
  ycl[0] = 1;
  for (int i = 1; i <= n; i++) ycl[i] = ycl[i - 1] * rev % mod;
  for (int i = 0; i < (1 << n); i++) {
    int sum = 0;
    for (int j = 0; j < n; j++)
      if (i >> j & 1) sum++;
    ans = (ans + a[i] * ycl[sum]) % mod;
  }
  cout << (long long)(ans + mod) % mod * 3 % mod << endl;
}
