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("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("popcnt")
#pragma GCC target("avx2")
using namespace std;
template <typename T1, typename T2>
inline void mine(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void maxe(T1 &x, T2 y) {
  if (x < y) x = y;
}
ostream &operator<<(ostream &a, const vector<int> &b) {
  for (auto k : b) cout << k << ' ';
  return a;
}
const int MOD = 1000000007;
const int INF = 1000000050;
const long long BIG = (long long)2e18 + 50;
const int MX = 200010;
const double EPS = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long add(long long a, long long b) {
  a += b;
  if (a >= MOD) return a - MOD;
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) return a + MOD;
  return a;
}
long long mult(long long a, long long b) { return a * b % MOD; }
long long powmod(long long a, long long p) {
  if (p == 0) return 1;
  if (p & 1) return mult(a, powmod(a, p - 1));
  return powmod(mult(a, a), p / 2);
}
long long rev(long long a) { return powmod(a, MOD - 2); }
void conv(vector<int> &a, int k) {
  for (int len = 1; len < (1 << k); len <<= 1) {
    for (int st = 0; st < (1 << k); st += (len << 1))
      for (int i = 0; i < len; i++) {
        long long x = a[st + i], y = a[st + len + i];
        a[st + i] = add(x, y);
        a[st + len + i] = sub(x, y);
      }
  }
}
void invconv(vector<int> &a, int k) {
  conv(a, k);
  int R = rev(1 << k);
  for (int &x : a) x = mult(x, R);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  string s;
  cin >> k >> s;
  n = 1 << k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) a[i] = s[i] == '1';
  conv(a, k);
  for (auto &x : a) x = mult(x, x);
  invconv(a, k);
  int ans = 0;
  for (int i = 0; i < n; i++)
    ans = add(ans, mult(a[i], 1 << (k - __builtin_popcount(i))));
  cout << mult(ans, 3) << '\n';
}
