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>
inline int in() {
  int k = 0;
  char ch = getchar();
  bool p = 1;
  while (ch < '-') ch = getchar();
  if (ch == '-') ch = getchar(), p = 0;
  while (ch > '-') k = k * 10 + ch - '0', ch = getchar();
  return p ? k : -k;
}
const int YL = 1e9 + 7, N = 1 << 20;
inline int MO(const int &x) { return x >= YL ? x - YL : x; }
inline int ksm(int a, int k) {
  int r = 1;
  while (k) {
    if (k & 1) r = 1ll * r * a % YL;
    a = 1ll * a * a % YL, k >>= 1;
  }
  return r;
}
int len, a[N], cnt[N];
void fwt(int *a, int opt = 1) {
  for (int st = 2, m = 1; st <= len; st <<= 1, m <<= 1)
    for (int *p = a; p != a + len; p += st)
      for (int k = 0, x, y; k < m; p[k] = MO(x + y), ++k)
        x = p[k], y = p[k + m], p[k + m] = MO(x - y + YL);
  if (opt == 1) return;
  int inv = ksm(len, YL - 2);
  for (int i = 0; i < len; ++i) a[i] = 1ll * a[i] * inv % YL;
}
char S[N];
int main() {
  int n = in();
  len = 1 << n;
  scanf("%s", S);
  for (int i = 0; i < len; ++i) a[i] = S[i] - '0';
  for (int i = 0; i < len; ++i) cnt[i] = cnt[i >> 1] + (i & 1);
  fwt(a);
  for (int i = 0; i < len; ++i) a[i] = 1ll * a[i] * a[i] % YL;
  fwt(a, -1);
  int ans = 0;
  for (int i = 0; i < len; ++i)
    ans = (1ll * a[i] * ksm(2, n - cnt[i]) + ans) % YL;
  ans = MO(MO(ans + ans) + ans);
  printf("%d\n", ans);
  return 0;
}
