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 = 0x3f3f3f3f, oo = inf;
inline long long read() {
  register long long x = 0;
  char f = 0;
  register char c = getchar();
  for (; !isdigit(c); c = getchar()) f |= (c == '-');
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return f ? -x : x;
}
inline double readdb() {
  register double x = 0, p = 0.1;
  register char f = 0, c = getchar();
  for (; !isdigit(c); c = getchar()) f |= (c == '-');
  for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
  if (c == '.')
    for (c = getchar(); isdigit(c); c = getchar(), p /= 10)
      x = x + (c ^ 48) * p;
  return f ? -x : x;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void writeln(long long x, char c, long long y) {
  write(x);
  putchar(c);
  writeln(y);
}
inline void writeln(long long x, char c, long long y, char d, long long z) {
  write(x);
  putchar(c);
  write(y);
  putchar(d);
  writeln(z);
}
const int Mod = 1e9 + 7;
const int maxn = (1 << 20) + 4;
int n, l;
long long ans;
int cnt[maxn];
long long a[maxn];
char s[maxn];
void FWT(long long a[], int n) {
  for (int i = 1; i < n; i <<= 1)
    for (int j = 0; j < n; j += (i << 1))
      for (int k = 0; k < i; ++k) {
        long long x = a[j + k], y = a[i + j + k];
        a[j + k] = (x + y);
        a[i + j + k] = (x - y);
      }
}
void IFWT(long long a[], int n) {
  for (int i = 1; i < n; i <<= 1)
    for (int j = 0; j < n; j += (i << 1))
      for (int k = 0; k < i; ++k) {
        long long x = a[j + k], y = a[i + j + k];
        a[j + k] = (x + y) / 2;
        a[i + j + k] = (x - y) / 2;
      }
}
int main() {
  n = read(), l = 1 << n;
  scanf("%s", s);
  for (register int i = (0); i < (l); ++i) a[i] = s[i] - '0';
  FWT(a, l);
  for (register int i = (0); i < (l); ++i) a[i] = a[i] * a[i];
  IFWT(a, l);
  ans = a[0] * l % Mod;
  for (int i = 1; i < l; i++)
    cnt[i] = cnt[i - (i & -i)] + 1,
    ans = (ans + a[i] * (1 << (n - cnt[i]))) % Mod;
  writeln(ans * 3 % Mod);
  return 0;
}
