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;
const int N = 100005;
const int P = 1e9 + 7;
const int inf = 1e9;
const long long Inf = 1e15;
inline int IN() {
  char ch = getchar();
  int x = 0, f = 0;
  while (ch < '0' || ch > '9') ch = getchar(), f = (ch == '-');
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - 48;
    ch = getchar();
  }
  return f ? (-x) : x;
}
inline int Pow(int x, int y, int p) {
  int an = 1;
  for (; y; y >>= 1, x = (long long)x * x % p)
    if (y & 1) an = (long long)an * x % p;
  return an;
}
void renew(int &x, int y) {
  x += y;
  if (x < 0)
    x += P;
  else if (x >= P)
    x -= P;
}
void setIO(string a) {
  freopen((a + ".in").c_str(), "r", stdin);
  freopen((a + ".out").c_str(), "w", stdout);
}
template <typename T>
inline void chkmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <typename T>
inline void chkmax(T &a, const T &b) {
  if (a < b) a = b;
}
int n, m;
char str[1 << 21];
int val[1 << 21], rev[1 << 21];
int gval[1 << 21];
void dft(int *a, int L, int R) {
  if (L + 1 == R) return;
  int md = (L + R) / 2;
  dft(a, L, md);
  dft(a, md, R);
  for (int i = (int)L; i < (int)md; i++) {
    int x = a[i], y = a[i - L + md];
    a[i] = (x + y) % P;
    a[i - L + md] = (x + P - y) % P;
  }
}
int ans = 0;
int clz[1 << 21];
int main() {
  scanf("%d", &n);
  m = 1 << n;
  scanf("%s", str + 1);
  for (int i = (int)0; i < (int)1 << n; i++)
    val[i] = str[i + 1] - '0', gval[i] = 1 - val[i];
  for (int i = (int)1; i < (int)1 << n; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (n - 1));
  dft(val, 0, m);
  dft(gval, 0, m);
  for (int i = (int)0; i < (int)m; i++)
    val[i] = (long long)val[i] * gval[i] % P;
  dft(val, 0, m);
  int rm = Pow(m, P - 2, P);
  for (int i = (int)0; i < (int)m; i++) val[i] = (long long)val[i] * rm % P;
  for (int i = (int)0; i < (int)m; i++) clz[i] = clz[i >> 1] + (i & 1);
  for (int i = (int)0; i < (int)m; i++)
    val[i] = (long long)val[i] * (1 << clz[i]) % P;
  for (int i = (int)0; i < (int)m; i++) ans = (ans + val[i]) % P;
  ans = (long long)ans * 3 % P;
  printf("%d\n", ans);
  return 0;
}
