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 ll = long long;
using namespace std;
using namespace std::chrono;
int n;
int const N = (1 << 20) + 3;
char a[N];
int sum_of_dig[N];
ll const P = 1e9 + 7;
vector<ll> f;
vector<ll> jej;
vector<ll> fft(const vector<ll>& inp) {
  if (inp.size() == 1) return inp;
  vector<ll> pom1;
  vector<ll> pom2;
  int k = (int)(inp.size()) / 2;
  for (int i = (0); i <= (k - 1); i++) {
    pom1.push_back(((inp[i] + inp[k + i]) * ((P + 1) / 2)) % P);
    pom2.push_back(((inp[i] - inp[k + i] + P) * ((P + 1) / 2)) % P);
  }
  pom1 = fft(pom1);
  pom2 = fft(pom2);
  for (int i = (0); i <= (k - 1); i++) pom1.push_back(pom2[i]);
  return pom1;
}
ll power(ll a, ll pow) {
  if (!pow) return 1;
  ll x = power(a, pow / 2);
  if (pow % 2)
    return (a * ((x * x) % P)) % P;
  else
    return (x * x) % P;
}
int main() {
  scanf("%d", &n);
  scanf("%s", &a);
  for (int i = (0); i <= ((1 << n) - 1); i++)
    f.push_back((a[i] == '1') ? 1 : 0);
  jej = fft(f);
  ll ans = 0;
  sum_of_dig[0] = 0;
  for (int i = (0); i <= ((1 << n) - 1); i++) {
    sum_of_dig[i] = (i % 2) ? sum_of_dig[i / 2] + 1 : sum_of_dig[i / 2];
    ans =
        (ans + (power(3, n - sum_of_dig[i]) * ((jej[i] * jej[i]) % P)) % P) % P;
  }
  ans = (ans * (3 * power(2, n)) % P) % P;
  printf("%I64d ", ans);
  return 0;
}
