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;
template <int n>
using AI = array<int, n>;
const double eps = 1e-9;
const int oo = 0x3F3F3F3F;
const long long ooLL = 0x3F3F3F3F3F3F3F3FLL;
const int MOD = 1000000007;
template <typename T>
T sqr(T x) {
  return x * x;
}
const double PI = acos(-1.);
long long Round(double x) { return x < 0 ? x - .5 : x + .5; }
int main() {
  void run();
  run();
  return 0;
}
int n, k;
long long c[1 << 20];
void tr() {
  for (int b = (0), _b(k); b < _b; ++b)
    for (int i = (0), _b(n); i < _b; ++i) {
      int j = i ^ (1 << b);
      if (j < i) continue;
      long long u = c[i], v = c[j];
      c[i] = u + v;
      c[j] = u - v;
    }
}
char s[1 << 20];
void run() {
  scanf("%d\n%s", &k, s);
  n = 1 << k;
  for (int i = (0), _b(n); i < _b; ++i) c[i] = s[i] - '0';
  tr();
  for (int i = (0), _b(n); i < _b; ++i) c[i] *= c[i];
  tr();
  for (int i = (0), _b(n); i < _b; ++i) c[i] /= n;
  long long rs = 0;
  for (int i = (0), _b(n); i < _b; ++i)
    rs = (rs + (1 << (k - __builtin_popcount(i))) * c[i]) % MOD;
  printf("%lld\n", rs * 3 % MOD);
}
