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 maxn = 15e5, mod = 1e9 + 7, modi = 5e8 + 4;
string s;
vector<int> v[2];
int n;
int arr1[maxn], arr2[maxn];
void transform(int *l, int *r, bool rev) {
  if (r - l == 1) return;
  int *mid = l + (r - l) / 2;
  transform(l, mid, rev);
  transform(mid, r, rev);
  for (int i = 0; i < (mid - l); i++) {
    int A = *(l + i), B = *(mid + i);
    if (rev) {
      *(l + i) = 1ll * (A + B) * modi % mod;
      *(mid + i) = 1ll * (A - B) * modi % mod;
    } else {
      *(l + i) = (A + B) % mod;
      *(mid + i) = (A - B) % mod;
    }
  }
}
int solve(int a, int b) {
  memset(arr1, 0, sizeof arr1);
  memset(arr2, 0, sizeof arr2);
  int N = (1 << n);
  for (int i = 0; i < (N / 2); i++) ++arr1[v[a][i]], ++arr2[v[b][i]];
  transform(arr1, arr1 + N, 0);
  transform(arr2, arr2 + N, 0);
  for (int i = 0; i < N; i++) arr1[i] = (1ll * arr1[i] * arr2[i]) % mod;
  transform(arr1, arr1 + N, 1);
  int bt = a ^ b, ans = 0;
  for (int i = 0; i < N; i++) {
    int c = __builtin_popcount(i);
    if (!bt) c = n - c;
    ans = (ans + 1ll * arr1[i] * (1 << c)) % mod;
  }
  return ans;
}
int main() {
  cin >> n >> s;
  for (int i = 0; i < int((s).size()); i++) v[(s[i] - '0')].push_back(i);
  long long ans = (1ll * solve(1, 1) + solve(0, 1) + solve(0, 0)) % mod;
  cout << ans << endl;
}
