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 INF = 1e9 + 7;
const long long INFL = 1e18 + 123;
const double PI = atan2(0, -1);
mt19937 tw(960172);
long long rnd(long long x, long long y) {
  static uniform_int_distribution<long long> d;
  return d(tw) % (y - x + 1) + x;
}
const int MAXN = 21;
const int MOD = 1e9 + 7;
char buf[1 << MAXN];
long long rsq[1 << MAXN];
int inp[1 << MAXN];
long long tmp[1 << MAXN];
int calc(int v, int vl, int vr) {
  if (vl + 1 == vr) {
    return 0;
  }
  int vm = (vl + vr) / 2;
  calc(v * 2, vl, vm);
  calc(v * 2 + 1, vm, vr);
  fill(tmp + vl, tmp + vr, 0);
  for (int i = vl; i < vr; ++i) {
    if (i < vm) {
      tmp[i] = (rsq[i] * 2 + rsq[i + vm - vl]) % MOD;
    } else {
      tmp[i] = (rsq[i] * 2 + rsq[i - (vm - vl)]) % MOD;
    }
  }
  for (int i = vl; i < vr; ++i) {
    rsq[i] = tmp[i];
  }
  return 0;
}
int main() {
  cerr << fixed << setprecision(15);
  cout << fixed << setprecision(15);
  int n;
  scanf("%d", &n);
  scanf("%s", buf);
  int bpv = 1 << n;
  for (int i = 0; i < bpv; ++i) {
    rsq[i] = buf[i] - '0';
    inp[i] = rsq[i];
  }
  calc(1, 0, bpv);
  long long ans = 0;
  for (int i = 0; i < bpv; ++i) {
    ans = (ans + rsq[i] * inp[i]) % MOD;
  }
  ans = (ans * 3) % MOD;
  cout << ans << "\n";
  return 0;
}
