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;
void per() { cerr << endl; }
template <typename Head, typename... Tail>
void per(Head H, Tail... T) {
  cerr << H << ' ';
  per(T...);
}
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
template <class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class U, class V>
istream& operator>>(istream& in, pair<U, V>& a) {
  return in >> a.first >> a.second;
}
template <typename W,
          typename T = typename enable_if<!is_same<W, string>::value,
                                          typename W::value_type>::type>
ostream& operator<<(ostream& out, const W& v) {
  out << "{ ";
  for (const auto& first : v) out << first << ", ";
  return out << '}';
}
template <class T>
void readArr(T from, T to) {
  for (auto i = from; i != to; ++i) cin >> *i;
}
mt19937 mrand(1337);
unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
unsigned long long myRand64() {
  return ((unsigned long long)myRand32() << 32) ^ myRand32();
}
const int mod = 1000000007;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void dec(int& a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mult(int a, int b) { return a * (long long)b % mod; }
int bp(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}
int n;
string s;
int K = 1 << 20;
void hadamard(vector<int>& v) {
  for (int step = K; step > 1; step /= 2) {
    for (int start = 0; start < K; start += step) {
      for (int w = 0; w < step / 2; w++) {
        int F = v[start + w];
        add(F, v[start + step / 2 + w]);
        int S = v[start + w];
        dec(S, v[start + step / 2 + w]);
        v[start + w] = F;
        v[start + step / 2 + w] = S;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> s;
  int res = 0;
  vector<int> f1(1 << n);
  vector<int> f2(1 << n);
  for (int m1 = 0; m1 < (1 << n); ++m1)
    if (s[m1] == '1') f1[m1] = 1;
  for (int m2 = 0; m2 < (1 << n); ++m2)
    if (s[m2] == '0') f2[m2] = 1;
  K = 1 << n;
  hadamard(f1);
  hadamard(f2);
  for (int i = 0; i < f1.size(); ++i) f1[i] = mult(f1[i], f2[i]);
  hadamard(f1);
  auto KEK = bp(K, mod - 2);
  for (int i = 0; i < f1.size(); ++i) f1[i] = mult(f1[i], KEK);
  for (int m1 = 0; m1 < (1 << n); ++m1)
    add(res, mult(f1[m1], 1 << __builtin_popcount(m1)));
  cout << mult(res, 3);
}
