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;
template <typename T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
const int INF = 0x3f3f3f3f;
const int N = 20, mod = 1e9 + 7, Inv = (mod + 1) / 2;
char str[1 << N];
int A[1 << N];
inline int read() {
  int x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') flag = -1, ch = getchar();
  while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch - '0'), ch = getchar();
  return x * flag;
}
inline int Mod(int val) {
  if (val >= mod) val -= mod;
  if (val < 0) val += mod;
  return val;
}
inline void FWT(int P[], int len, int type) {
  for (int i = 2; i <= len; i <<= 1)
    for (int j = 0; j < len; j += i)
      for (int k = (0), kend = ((i >> 1) - 1); k <= kend; k++) {
        int x = P[j + k], y = P[j + k + (i >> 1)];
        if (type)
          P[j + k] = Mod(x + y), P[j + k + (i >> 1)] = Mod(x - y);
        else
          P[j + k] = 1ll * (x + y) * Inv % mod,
                P[j + k + (i >> 1)] = 1ll * Mod(x - y) * Inv % mod;
      }
}
int main() {
  int n = read(), ans = 0;
  scanf("%s", str);
  for (int i = (0), iend = ((1 << n) - 1); i <= iend; i++) A[i] = str[i] - '0';
  FWT(A, 1 << n, 1);
  for (int i = (0), iend = ((1 << n) - 1); i <= iend; i++)
    A[i] = 1ll * A[i] * A[i] % mod;
  FWT(A, 1 << n, 0);
  for (int i = (0), iend = ((1 << n) - 1); i <= iend; i++)
    ans = (ans + (1ll << (n - __builtin_popcount(i))) * A[i]) % mod;
  printf("%lld\n", ans * 3ll % mod);
  return 0;
}
