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>
template <typename T>
inline void rd(T& x) {
  int si = 1;
  char c = getchar();
  x = 0;
  while (!isdigit(c)) si = c == '-' ? -1 : si, c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  x *= si;
}
template <typename T, typename... Args>
inline void rd(T& x, Args&... args) {
  rd(x);
  rd(args...);
}
const int kN = 2e6 + 5, kMod = 1e9 + 7;
int QPow(int a, int b) {
  int ret = 1, bas = a;
  for (; b; b >>= 1, bas = 1LL * bas * bas % kMod)
    if (b & 1) ret = 1LL * ret * bas % kMod;
  return ret;
}
void FWTxnor(int f[], int len, bool flag) {
  int x = flag ? 1 : QPow(2, kMod - 2);
  for (int i = 1; i < 1 << len; i <<= 1)
    for (int j = 0; j < 1 << len; j += i << 1)
      for (int k = 0; k < i; ++k) {
        int l = f[j + k], r = f[i + j + k];
        f[j + k] = 1LL * (l + r) % kMod * x % kMod;
        f[i + j + k] = 1LL * (l - r + kMod) % kMod * x % kMod;
      }
}
int n, s[kN];
char str[kN];
int main() {
  int T = 1;
  while (T--) {
    int ans = 0;
    rd(n);
    scanf("%s", str);
    for (int i = 0; i < 1 << n; ++i) s[i] = str[i] - '0';
    FWTxnor(s, n, true);
    for (int i = 0; i < 1 << n; ++i) s[i] = 1LL * s[i] * s[i] % kMod;
    FWTxnor(s, n, false);
    for (int i = 0; i < 1 << n; ++i) {
      int pcnt = __builtin_popcount(i);
      ans = (ans + (1LL << (n - pcnt)) * s[i]) % kMod;
    }
    ans = 3LL * ans % kMod;
    printf("%d\n", ans);
  }
  return 0;
}
