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 mod = 1e9 + 7;
long long powr(long long a, long long b) {
  long long x = 1;
  for (; b; b >>= 1, a = (a * 1ll * a) % mod)
    if (b & 1) x = (x * a) % mod;
  return x;
}
long long inv(long long a, int md) { return powr(a, md - 2); }
vector<int> xor_convolution(vector<int>& a, vector<int>& b, int k) {
  if (k <= 4) {
    vector<int> C(1 << k, 0);
    for (__typeof(1 << k) i = (0) - ((0) > (1 << k));
         i != (1 << k) - ((0) > (1 << k)); i += 1)
      for (__typeof(1 << k) j = (0) - ((0) > (1 << k));
           j != (1 << k) - ((0) > (1 << k)); j += 1) {
        C[i ^ j] = (C[i ^ j] + a[i] * 1ll * b[j]) % mod;
      }
    return C;
  }
  int inv2 = inv(2, mod);
  vector<int> a1(1 << k);
  vector<int> A(a.begin(), a.begin() + (1 << (k - 1))),
      B(a.begin() + (1 << (k - 1)), a.end());
  vector<int> C(b.begin(), b.begin() + (1 << (k - 1))),
      D(b.begin() + (1 << (k - 1)), b.end());
  vector<int> X(1 << (k - 1)), Y(1 << (k - 1)), P(1 << (k - 1)),
      Q(1 << (k - 1));
  for (__typeof(1 << (k - 1)) i = (0) - ((0) > (1 << (k - 1)));
       i != (1 << (k - 1)) - ((0) > (1 << (k - 1))); i += 1) {
    X[i] = A[i] + B[i], Y[i] = C[i] + D[i];
    if (X[i] >= mod) X[i] -= mod;
    if (Y[i] >= mod) Y[i] -= mod;
  }
  for (__typeof(1 << (k - 1)) i = (0) - ((0) > (1 << (k - 1)));
       i != (1 << (k - 1)) - ((0) > (1 << (k - 1))); i += 1) {
    P[i] = A[i] - B[i], Q[i] = C[i] - D[i];
    if (P[i] < 0) P[i] += mod;
    if (Q[i] < 0) Q[i] += mod;
  }
  vector<int> Z1 = xor_convolution(P, Q, k - 1);
  vector<int> Z2 = xor_convolution(X, Y, k - 1);
  int hf = 1 << (k - 1);
  for (__typeof(hf) i = (0) - ((0) > (hf)); i != (hf) - ((0) > (hf)); i += 1) {
    a1[i] = Z1[i] + Z2[i];
    if (a1[i] >= mod) a1[i] -= mod;
    a1[i] = (a1[i] * 1ll * inv2) % mod;
  }
  for (__typeof(hf + hf) i = (hf) - ((hf) > (hf + hf));
       i != (hf + hf) - ((hf) > (hf + hf)); i += 1) {
    a1[i] = Z2[i - hf] - Z1[i - hf];
    if (a1[i] < 0) a1[i] += mod;
    a1[i] = (a1[i] * 1ll * inv2) % mod;
  }
  return a1;
}
vector<int> naive_xor_mul(vector<int>& A, vector<int>& B) {
  int x = max(int((A).size()), int((B).size()));
  int _sz = 1, k = 0;
  while (_sz < x) _sz <<= 1, k++;
  A.resize(_sz), B.resize(_sz);
  vector<int> C(_sz, 0);
  for (__typeof(_sz) i = (0) - ((0) > (_sz)); i != (_sz) - ((0) > (_sz));
       i += 1)
    for (__typeof(_sz) j = (0) - ((0) > (_sz)); j != (_sz) - ((0) > (_sz));
         j += 1)
      C[i ^ j] = (C[i ^ j] + A[i] * 1ll * B[j]) % mod;
  return C;
}
vector<int> xor_mul(vector<int>& A, vector<int>& B) {
  int x = max(int((A).size()), int((B).size()));
  int _sz = 1, k = 0;
  while (_sz < x) _sz <<= 1, k++;
  A.resize(_sz), B.resize(_sz);
  return xor_convolution(A, B, k);
}
vector<int> A, B;
const int N = 2e6 + 12;
int pw2[N];
char s[N];
int main() {
  pw2[0] = 1;
  for (int i = 1; i < N; i++) pw2[i] = (pw2[i - 1] << 1) % mod;
  int n;
  scanf("%d ", &n);
  scanf("%s", s);
  A.resize(1 << n);
  B.resize(1 << n);
  for (int i = 0; i < (1 << n); i++) {
    A[i] = B[i] = s[i] - '0';
  }
  A = xor_mul(A, B);
  int ans = 0;
  for (int i = 0; i < (1 << n); i++) {
    int k = __builtin_popcount(i);
    ans = (ans + pw2[n - k] * 1ll * A[i]) % mod;
  }
  ans = (ans * 1ll * 3) % mod;
  printf("%d\n", ans);
}
