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 double pi = acos(-1.0);
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int alf = 26;
const int dig = 10;
const int two = 2;
const int th = 3;
const long long prost = 239;
const long long btc = 30;
const double eps = 1e-8;
const long long INF = (long long)(1e18 + 239);
const int BIG = (int)(1e9 + 239);
const long long MOD = (int)(1e9 + 7);
const long long MOD2 = (long long)MOD * (long long)MOD;
mt19937_64 rnd(239);
const int M = (int)(2e5 + 239);
const int N = (int)(2e3 + 239);
const int L = 18;
const int T = (1 << 19);
const int B = 500;
const int X = (1 << 20) + 239;
int power(int a, int k) {
  if (k == 0) return 1;
  int t = power(a, k >> 1);
  t = 1LL * t * t % MOD;
  if (k & 1) t = 1LL * t * a % MOD;
  return t;
}
string s;
int n, pw2[X];
void hadamart(vector<int>& b) {
  for (int t = 1; t < (1 << n); t *= 2)
    for (int s = 0; s < (1 << n); s += 2 * t)
      for (int i = 0; i < t; i++) {
        int x = b[s + i];
        int y = b[s + t + i];
        b[s + i] = x + y;
        if (b[s + i] >= MOD) b[s + i] -= MOD;
        b[s + t + i] = x - y;
        if (b[s + t + i] < 0) b[s + t + i] += MOD;
      }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pw2[0] = 1;
  for (int i = 1; i < X; i++) {
    pw2[i] = pw2[i - 1] * 2;
    if (pw2[i] >= MOD) pw2[i] -= MOD;
  }
  cin >> n >> s;
  vector<int> a((1 << n), 0);
  for (int i = 0; i < (1 << n); i++) a[i] = (s[i] == '1');
  vector<int> b((1 << n), 0);
  for (int i = 0; i < (1 << n); i++) b[i] = (s[i] == '0');
  hadamart(a);
  hadamart(b);
  for (int i = 0; i < (1 << n); i++) a[i] = 1LL * a[i] * b[i] % MOD;
  hadamart(a);
  int u = power((1 << n), MOD - 2);
  for (int i = 0; i < (1 << n); i++) a[i] = 1LL * a[i] * u % MOD;
  int ans = 0;
  for (int i = 0; i < (1 << n); i++) {
    ans += 1LL * a[i] * pw2[__builtin_popcount(i)] % MOD;
    if (ans >= MOD) ans -= MOD;
  }
  cout << (3LL * ans % MOD);
  return 0;
}
