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 = (1 << 20);
const long long MOD = 1e9 + 7;
const long long INV2 = 5e8 + 4;
template <typename T>
inline void read(T &AKNOI) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  AKNOI = flag * x;
}
char s[MAXN];
int n, len, bit[MAXN];
long long a[MAXN], pw2[21], ans;
void FWT(long long *a, int len, int type) {
  for (int i = 1; i < len; i <<= 1) {
    int p = (i << 1);
    for (int j = 0; j < len; j += p) {
      for (int k = 0; k < i; ++k) {
        long long x = a[j + k], y = a[j + k + i];
        a[j + k] = (x + y) % MOD;
        a[j + k + i] = (x - y + MOD) % MOD;
        if (type == -1) {
          a[j + k] = a[j + k] * INV2 % MOD;
          a[j + k + i] = a[j + k + i] * INV2 % MOD;
        }
      }
    }
  }
}
void init() {
  read(n);
  len = (1 << n);
  scanf("%s", s);
  pw2[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw2[i] = pw2[i - 1] * 2 % MOD;
  }
  for (int i = 0; i < len; ++i) {
    bit[i] = bit[i >> 1] + (i & 1);
    a[i] = s[i] - '0';
  }
}
void solve() {
  FWT(a, len, 1);
  for (int i = 0; i < len; ++i) {
    a[i] = a[i] * a[i] % MOD;
  }
  FWT(a, len, -1);
  for (int i = 0; i < len; ++i) {
    ans = (ans + a[i] * pw2[n - bit[i]]) % MOD;
  }
  printf("%lld\n", ans * 3 % MOD);
}
int main() {
  init();
  solve();
  return 0;
}
