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 mod = 1000000007;
const int inv2 = (mod + 1) / 2;
const int MAXN = (1 << 20);
void fft(vector<long long> &a, int low, int high) {
  if (low == high - 1) return;
  int len = (high - low) / 2, mid = low + len;
  fft(a, low, mid);
  fft(a, mid, high);
  for (int i = low; i < mid; i++) {
    auto x1 = a[i];
    auto x2 = a[i + len];
    a[i] = (x1 - x2) % mod;
    a[i + len] = (x1 + x2) % mod;
  }
}
void inv_fft(vector<long long> &a, int low, int high) {
  if (low == high - 1) return;
  int len = (high - low) / 2, mid = low + len;
  for (int i = low; i < mid; i++) {
    auto y1 = a[i];
    auto y2 = a[i + len];
    a[i] = (y1 + y2) * inv2 % mod;
    a[i + len] = (y2 - y1) * inv2 % mod;
  }
  inv_fft(a, low, mid);
  inv_fft(a, mid, high);
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int n;
  cin >> n;
  vector<int> popc(1 << n);
  for (int i = 1; i < popc.size(); ++i) {
    popc[i] = popc[i / 2] + i % 2;
  }
  string s;
  cin >> s;
  long long res = 0;
  vector<long long> A(1 << n);
  for (int i = 0; i < s.size(); ++i) {
    s[i] -= '0';
    A[i] = s[i];
  }
  fft(A, 0, A.size());
  for (int i = 0; i < A.size(); i++) A[i] = A[i] * A[i] % mod;
  inv_fft(A, 0, A.size());
  for (int i = 0; i < A.size(); ++i) {
    res = (res + A[i] % mod * (1LL << (n - popc[i]))) % mod;
  }
  cout << 3 * (res % mod + mod) % mod << endl;
  return 0;
}
