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 void gmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <class T>
inline void gmax(T &x, const T &y) {
  if (x < y) x = y;
}
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Bufferhead, *Buffertail;
bool Terminal;
inline char Getchar() {
  if (Bufferhead == Buffertail) {
    int l = fread(buffer, 1, BufferSize, stdin);
    if (!l) {
      Terminal = 1;
      return 0;
    }
    Buffertail = (Bufferhead = buffer) + l;
  }
  return *Bufferhead++;
}
template <class T>
inline bool read(T &x) {
  x = 0;
  char c = Getchar(), rev = 0;
  while (c < '0' || c > '9') {
    c = Getchar();
    rev |= c == '-';
    if (Terminal) return 0;
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = Getchar();
  if (c == '.') {
    c = Getchar();
    double t = 0.1;
    while (c >= '0' && c <= '9')
      x = x + (c - '0') * t, c = Getchar(), t = t / 10;
  }
  x = rev ? -x : x;
  return 1;
}
template <class T1, class T2>
inline bool read(T1 &x, T2 &y) {
  return read(x) & read(y);
}
template <class T1, class T2, class T3>
inline bool read(T1 &x, T2 &y, T3 &z) {
  return read(x) & read(y) & read(z);
}
template <class T1, class T2, class T3, class T4>
inline bool read(T1 &x, T2 &y, T3 &z, T4 &w) {
  return read(x) & read(y) & read(z) & read(w);
}
inline bool reads(long long *x) {
  char c = Getchar();
  while (c < 33 || c > 126) {
    c = Getchar();
    if (Terminal) return 0;
  }
  while (c >= 33 && c <= 126) (*x++) = c - '0', c = Getchar();
  *x = 0;
  return 1;
}
template <class T>
inline void print(T x, const char c = '\n') {
  if (!x) {
    putchar('0');
    putchar(c);
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  int m = 0, a[20];
  while (x) a[m++] = x % 10, x /= 10;
  while (m--) putchar(a[m] + '0');
  putchar(c);
}
const int inf = 0x3f3f3f3f;
const int N = 1100005, M = 100005, mod = 1e9 + 7;
template <class T, class S>
inline void ch(T &x, const S y) {
  x = (x + y) % mod;
}
inline int exp(int x, int y, const int mod = ::mod) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = (long long)ans * x % mod;
    x = (long long)x * x % mod;
    y >>= 1;
  }
  return ans;
}
int n;
char bits[1 << 20];
long long a[1 << 20];
void fwt(long long *a, int l = 0, int r = (1 << n) - 1) {
  if (l == r) return;
  int m = (1 << n);
  for (int i = 2; i <= m; i <<= 1) {
    int c = i >> 1;
    for (int j = 0; j < m; j += i)
      for (int k = 0; k < c; k++) {
        long long y = a[j + k + c];
        a[j + k + c] = a[j + k] - y;
        a[j + k] += y;
      }
  }
}
int main() {
  read(n);
  reads(a);
  for (int i = 1; i < (1 << n); i++) bits[i] = bits[i ^ (i & -i)] + 1;
  fwt(a);
  for (int i = 0; i < (1 << n); i++) a[i] = ((a[i]) * (a[i]));
  fwt(a);
  long long ans = 0;
  for (int i = 0; i < (1 << n); i++) ans += a[i] >> bits[i];
  ans = ans * 3;
  print(ans % mod);
  return 0;
}
