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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
void FWHT(std::vector<lli>& P, bool inverse) {
  lli u, v;
  for (int len = 1; 2 * len <= ((lli)(P).size()); len <<= 1) {
    for (int i = 0; i < ((lli)(P).size()); i += 2 * len) {
      for (int j = 0; j < len; j++) {
        u = P[i + j];
        v = P[i + len + j];
        P[i + j] = u + v;
        P[i + len + j] = u - v;
      }
    }
  }
  if (inverse) {
    for (int i = 0; i < ((lli)(P).size()); i++) P[i] = P[i] / ((lli)(P).size());
  }
}
int tejas_919(int kkkk) {
  lli n, m, k, q, u, v, temp = 0, ans = 0;
  cin >> m;
  string s;
  cin >> s;
  n = 1LL << m;
  lli po[22];
  for (int i = 0; i < 22; i++) po[i] = 1LL << i;
  vl a(n);
  for (int i = 0; i < n; i++) a[i] = s[i] - '0';
  FWHT(a, 0);
  for (int i = 0; i < n; i++) a[i] *= a[i];
  FWHT(a, 1);
  for (int i = 0; i < n; i++) a[i] %= 1000000007;
  for (int i = 0; i < n; i++)
    ans += (a[i] * po[m - __builtin_popcountll((ulli)i)]) % 1000000007;
  cout << (ans * 3LL) % 1000000007 << '\n';
  return 0;
}
signed main() {
  if (!0) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    tejas_919(i + 1);
  }
  {};
  if (0) system("pause");
}
