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;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
struct modint {
  int x;
  modint(int o = 0) { x = o; }
  modint &operator=(int o) { return x = o, *this; }
  modint &operator+=(modint o) {
    return x = x + o.x >= 1000000007 ? x + o.x - 1000000007 : x + o.x, *this;
  }
  modint &operator-=(modint o) {
    return x = x - o.x < 0 ? x - o.x + 1000000007 : x - o.x, *this;
  }
  modint &operator*=(modint o) { return x = 1ll * x * o.x % 1000000007, *this; }
  modint &operator^=(int b) {
    modint a = *this, c = 1;
    for (; b; b >>= 1, a *= a)
      if (b & 1) c *= a;
    return x = c.x, *this;
  }
  modint &operator/=(modint o) { return *this *= o ^= 1000000007 - 2; }
  modint &operator+=(int o) {
    return x = x + o >= 1000000007 ? x + o - 1000000007 : x + o, *this;
  }
  modint &operator-=(int o) {
    return x = x - o < 0 ? x - o + 1000000007 : x - o, *this;
  }
  modint &operator*=(int o) { return x = 1ll * x * o % 1000000007, *this; }
  modint &operator/=(int o) { return *this *= ((modint(o)) ^= 1000000007 - 2); }
  template <class I>
  friend modint operator+(modint a, I b) {
    return a += b;
  }
  template <class I>
  friend modint operator-(modint a, I b) {
    return a -= b;
  }
  template <class I>
  friend modint operator*(modint a, I b) {
    return a *= b;
  }
  template <class I>
  friend modint operator/(modint a, I b) {
    return a /= b;
  }
  friend modint operator^(modint a, int b) { return a ^= b; }
  friend bool operator==(modint a, int b) { return a.x == b; }
  friend bool operator!=(modint a, int b) { return a.x != b; }
  bool operator!() { return !x; }
  modint operator-() { return x ? 1000000007 - x : 0; }
};
int n, mxs;
modint res, f[(1 << 20)];
inline void fwtand(modint *f, int n, int x) {
  for (int l = 2, k = 1; l <= n; l <<= 1, k <<= 1)
    for (int i = 0; i < n; i += l)
      for (int j = 0; j < k; ++j) f[i + j] = f[i + j] + f[i + j + k] * x;
}
inline void fwtor(modint *f, int n, int x) {
  for (int l = 2, k = 1; l <= n; l <<= 1, k <<= 1)
    for (int i = 0; i < n; i += l)
      for (int j = 0; j < k; ++j) f[i + j + k] = f[i + j + k] + f[i + j] * x;
}
inline void fwtxor(modint *f, int n, int x) {
  for (int l = 2, k = 1; l <= n; l <<= 1, k <<= 1)
    for (int i = 0; i < n; i += l)
      for (int j = 0; j < k; ++j) {
        modint a = f[i + j], b = f[i + j + k];
        f[i + j] = (a + b) * x, f[i + j + k] = (a - b) * x;
      }
}
int cnt[(1 << 20)];
char str[(1 << 20)];
signed main() {
  n = read(), mxs = (1 << n);
  scanf("%s", str);
  for (register int i = (0); i <= (mxs - 1); ++i) f[i] = (str[i] - '0');
  fwtxor(f, mxs, 1);
  for (register int i = (0); i <= (mxs - 1); ++i) f[i] = f[i] * f[i];
  fwtxor(f, mxs, 500000004);
  for (register int i = (0); i <= (mxs - 1); ++i) {
    if (i) cnt[i] = cnt[i >> 1] + (i & 1);
    res += f[i] * (modint(2) ^ (n - cnt[i]));
  }
  res *= 3;
  cout << res.x;
  return 0;
}
