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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
int IT_MAX = 1 << 15;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
vector<long long> mymul(vector<long long> A, vector<long long> B) {
  if (A.size() == 1 && B.size() == 1) {
    vector<long long> rv(1);
    rv[0] = A[0] * B[0];
    return rv;
  }
  int N = A.size(), i, j;
  vector<long long> Vu1;
  vector<long long> Vu2;
  vector<long long> rv1;
  vector<long long> rv2;
  for (i = 0; i < N / 2; i++) Vu1.push_back(A[i] + A[i + N / 2]);
  for (i = 0; i < N / 2; i++) Vu2.push_back(B[i] + B[i + N / 2]);
  rv1 = mymul(Vu1, Vu2);
  Vu1.clear();
  Vu2.clear();
  for (i = 0; i < N / 2; i++) Vu1.push_back(A[i] - A[i + N / 2]);
  for (i = 0; i < N / 2; i++) Vu2.push_back(B[i] - B[i + N / 2]);
  rv2 = mymul(Vu1, Vu2);
  vector<long long> rv;
  for (i = 0; i < N / 2; i++) rv.push_back((rv1[i] + rv2[i]) / 2);
  for (i = 0; i < N / 2; i++) rv.push_back((rv1[i] - rv2[i]) / 2);
  return rv;
}
char in[1100000];
int main() {
  vector<long long> Vu;
  int N, i, j;
  scanf("%d", &N);
  scanf("%s", in);
  for (i = 0; i < (1 << N); i++) Vu.push_back(in[i] - '0');
  vector<long long> rv = mymul(Vu, Vu);
  long long ans = 0;
  for (i = 0; i < (1 << N); i++) {
    long long v = rv[i];
    for (j = 0; j < N; j++)
      if (!(i & (1 << j))) v = (v * 2) % MOD;
    ans += v;
    ans %= MOD;
  }
  return !printf("%lld\n", ans * 3 % MOD);
}
