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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("sse,sse2,ssse3,sse3,sse4,popcnt,avx,mmx,abm,tune=native")
using namespace std;
namespace output {
void __(short x) { cout << x; }
void __(unsigned x) { cout << x; }
void __(int x) { cout << x; }
void __(long long x) { cout << x; }
void __(unsigned long long x) { cout << x; }
void __(double x) { cout << x; }
void __(long double x) { cout << x; }
void __(char x) { cout << x; }
void __(const char* x) { cout << x; }
void __(const string& x) { cout << x; }
void __(bool x) { cout << (x ? "true" : "false"); }
template <class S, class T>
void __(const pair<S, T>& x) {
  __(1 ? "(" : ""), __(x.first), __(1 ? ", " : " "), __(x.second),
      __(1 ? ")" : "");
}
template <class T>
void __(const vector<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class T>
void __(const set<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class T>
void __(const multiset<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class S, class T>
void __(const map<S, T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
void pr() { cout << "\n"; }
template <class S, class... T>
void pr(const S& a, const T&... b) {
  __(a);
  if (sizeof...(b)) __(' ');
  pr(b...);
}
}  // namespace output
using namespace output;
const int MN = 2e6 + 6, mod = 1e9 + 7;
int n, i, j, f[MN], ans, c1, c2, pw[22] = {1};
int qpow(int b, int exp) {
  if (exp <= 0)
    return 1;
  else if (exp & 1)
    return 1LL * b * qpow(b, exp - 1) % mod;
  else {
    int r = qpow(b, exp >> 1);
    return 1LL * r * r % mod;
  }
}
inline int popcnt(int x) {
  int res = 0;
  while (x) x -= x & -x, res++;
  return res;
}
void transform(int l, int r) {
  if (l == r) return;
  int m = (l + r) >> 1;
  transform(l, m);
  transform(m + 1, r);
  for (int i = 0; i <= m - l; i++) {
    int a = f[l + i];
    int b = f[m + 1 + i];
    f[l + i] = a + b;
    if (f[l + i] >= mod) f[l + i] -= mod;
    f[m + 1 + i] = a - b;
    if (f[m + 1 + i] < 0) f[m + 1 + i] += mod;
  }
}
int main() {
  scanf("%d\n", &n);
  for (i = 0; i < (1 << n); i++) f[i] = getchar() == '1';
  transform(0, (1 << n) - 1);
  for (i = 1; i <= n; i++) pw[i] = pw[i - 1] * 2;
  for (i = 0; i < (1 << n); i++) f[i] = (1LL * f[i] * f[i]) % mod;
  transform(0, (1 << n) - 1);
  int inv = qpow(qpow(2, mod - 2), n);
  for (i = 0; i < (1 << n); i++) f[i] = 1LL * f[i] * inv % mod;
  for (i = 0; i < (1 << n); i++)
    ans = (ans + 1LL * pw[popcnt(i ^ ((1 << n) - 1))] * f[i]) % mod;
  printf("%d\n", 3LL * ans % mod);
  return 0;
}
