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 <class T>
inline T Tmin(const T &lch, const T &rch) {
  if (lch < rch) return lch;
  return rch;
}
template <class T>
inline T Tmax(const T &lch, const T &rch) {
  if (lch < rch) return rch;
  return lch;
}
const int Inf = 0x3f3f3f3f;
const int Mod = 1e9 + 7;
const int Gp = 3;
inline int R_int() {
  register int n = 0;
  register char ch = getchar();
  register bool I = false;
  while (ch < '0' || ch > '9') I = (ch == '-' ? 1 : 0), ch = getchar();
  while (ch >= '0' && ch <= '9')
    n = (n << 1) + (n << 3) + (ch ^ '0'), ch = getchar();
  return I ? -n : n;
}
inline int Add(int a, int b) {
  int ans = (a + b);
  ans = (ans >= Mod ? ans - Mod : ans);
  return ans;
}
inline int Mis(int a, int b) {
  int ans = a - b;
  ans = (ans < 0 ? ans + Mod : ans);
  return ans;
}
inline int Mul(int a, int b) { return a * 1ll * b % Mod; }
inline int Pow(int a, int k) {
  int ans = 1;
  while (k) {
    if (k & 1) ans = Mul(ans, a);
    a = Mul(a, a);
    k = k >> 1;
  }
  return ans;
}
const int Inv2 = Pow(2, Mod - 2);
const int maxn = (1 << 21);
int bit[maxn];
int a[maxn];
int ans;
int n;
inline void FWT(int *a, int n, int opt) {
  for (int i = 1, x, y; i < n; i = i << 1) {
    for (int j = 0; j < n; j += i + i) {
      for (int k = 0; k < i; k++) {
        x = a[j + k];
        y = a[j + k + i];
        a[j + k] = Add(x, y);
        a[j + k + i] = Mis(x, y);
      }
    }
  }
  if (opt == -1) {
    int I = Pow(n, Mod - 2);
    for (int i = 0; i < n; i++) a[i] = Mul(a[i], I);
  }
}
signed main() {
  n = R_int();
  for (int i = 0; i < (1 << n); i++) a[i] = getchar() - '0';
  FWT(a, 1 << n, 1);
  for (int i = 0; i < (1 << n); i++) a[i] = Mul(a[i], a[i]);
  FWT(a, 1 << n, -1);
  for (int i = 0; i < (1 << n); i++) bit[i] = (bit[i >> 1]) + (i & 1);
  for (int i = 0; i < (1 << n); i++)
    ans = Add(ans, Mul(a[i], Pow(2, n - bit[i])));
  printf("%d\n", Mul(ans, 3));
  return 0;
}
