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 = 2e6 + 5;
const int mod = 1e9 + 7;
const int I2 = (mod + 1) / 2;
namespace {
int add(int x) { return (x >= mod) ? x - mod : x; }
int sub(int x) { return (x < 0) ? x + mod : x; }
void Add(int &x, int y) { x = add(x + y); }
void Sub(int &x, int y) { x = sub(x - y); }
int Pow(int x, int y = mod - 2) {
  int res = 1;
  for (; y; y >>= 1, x = 1LL * x * x % mod) {
    if (y & 1) res = 1LL * res * x % mod;
  }
  return res;
}
}  // namespace
int n, m, x, y, dp[N], B[N];
char s[N];
int main() {
  cin >> n;
  scanf("%s", s);
  for (int i = 0; i < (1 << n); ++i) dp[i] = s[i] - '0';
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < (1 << n); j += (1 << (i + 1)))
      for (int k = j; k < j + (1 << i); ++k) {
        int L = dp[k], R = dp[k + (1 << i)];
        dp[k] = add(L + R);
        dp[k + (1 << i)] = sub(L - R);
      }
  }
  for (int i = 0; i < (1 << n); ++i) dp[i] = 1LL * dp[i] * dp[i] % mod;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < (1 << n); j += (1 << (i + 1)))
      for (int k = j; k < j + (1 << i); ++k) {
        int L = dp[k], R = dp[k + (1 << i)];
        dp[k] = 1LL * I2 * add(L + R) % mod;
        dp[k + (1 << i)] = 1LL * I2 * sub(L - R) % mod;
      }
  }
  int ans = 0;
  for (int i = 0; i < (1 << n); ++i) {
    B[i] = (B[i >> 1]) + (i & 1);
    Add(ans, (1LL << (n - B[i])) * dp[i] % mod);
  }
  cout << 3LL * ans % mod << endl;
}
