#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 20, MOD = 1e9 + 7, INV2 = MOD + 1 >> 1;
inline int add(int a, const int &b) { return (a += b) >= MOD ? a - MOD : a; }
inline int sub(int a, const int &b) { return (a -= b) < 0 ? a + MOD : a; }
inline int mul(const int &a, const int &b) { return 1ll * a * b % MOD; }
inline int &inc(int &a, const int &b) { return a = add(a, b); }
inline int &dec(int &a, const int &b) { return a = sub(a, b); }
inline int &pro(int &a, const int &b) { return a = mul(a, b); }
inline int qpow(int a, int b) {
  int c = 1;
  for (; b; b >>= 1, pro(a, a))
    if (b & 1) pro(c, a);
  return c;
}
int a[N];
int n;
inline void fwt(int *f, int l, int opt) {
  for (int i = 1; i < l; i <<= 1)
    for (int j = 0; j < l; j += i << 1)
      for (int k = 0; k < i; k++) {
        int x = f[j + k], y = f[i + j + k];
        f[j + k] = add(x, y);
        f[i + j + k] = sub(x, y);
        if (opt) pro(f[j + k], INV2), pro(f[i + j + k], INV2);
      }
}
char s[N + 5];
int main() {
  scanf("%d%s", &n, s);
  for (int i = 0; i < 1 << n; i++) a[i] = s[i] - '0';
  fwt(a, 1 << n, 0);
  for (int i = 0; i < 1 << n; i++) pro(a[i], a[i]);
  fwt(a, 1 << n, 1);
  int ans = 0;
  for (int i = 0; i < 1 << n; i++)
    inc(ans, mul(a[i], qpow(2, n - __builtin_popcount(i))));
  printf("%d\n", mul(ans, 3));
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 20, MOD = 1e9 + 7, INV2 = MOD + 1 >> 1;
inline int add(int a, const int &b) { return (a += b) >= MOD ? a - MOD : a; }
inline int sub(int a, const int &b) { return (a -= b) < 0 ? a + MOD : a; }
inline int mul(const int &a, const int &b) { return 1ll * a * b % MOD; }
inline int &inc(int &a, const int &b) { return a = add(a, b); }
inline int &dec(int &a, const int &b) { return a = sub(a, b); }
inline int &pro(int &a, const int &b) { return a = mul(a, b); }
inline int qpow(int a, int b) {
  int c = 1;
  for (; b; b >>= 1, pro(a, a))
    if (b & 1) pro(c, a);
  return c;
}
int a[N];
int n;
inline void fwt(int *f, int l, int opt) {
  for (int i = 1; i < l; i <<= 1)
    for (int j = 0; j < l; j += i << 1)
      for (int k = 0; k < i; k++) {
        int x = f[j + k], y = f[i + j + k];
        f[j + k] = add(x, y);
        f[i + j + k] = sub(x, y);
        if (opt) pro(f[j + k], INV2), pro(f[i + j + k], INV2);
      }
}
char s[N + 5];
int main() {
  scanf("%d%s", &n, s);
  for (int i = 0; i < 1 << n; i++) a[i] = s[i] - '0';
  fwt(a, 1 << n, 0);
  for (int i = 0; i < 1 << n; i++) pro(a[i], a[i]);
  fwt(a, 1 << n, 1);
  int ans = 0;
  for (int i = 0; i < 1 << n; i++)
    inc(ans, mul(a[i], qpow(2, n - __builtin_popcount(i))));
  printf("%d\n", mul(ans, 3));
  return 0;
}

```