Problem: Toad Mikhail has an array of 2^k integers a_1, a_2, …, a_{2^k}.

Find two permutations p and q of integers 0, 1, …, 2^k-1, such that a_i is equal to p_i ⊕ q_i for all possible i, or determine there are no such permutations. Here ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Input

The first line contains one integer k (2 ≤ k ≤ 12), denoting that the size of the array is 2^k.

The next line contains 2^k space-separated integers a_1, a_2, …, a_{2^k} (0 ≤ a_i < 2^k) — the elements of the given array.

Output

If the given array can't be represented as element-wise XOR of two permutations of integers 0, 1, …, 2^k-1, print "Fou".

Otherwise, print "Shi" in the first line.

The next two lines should contain the description of two suitable permutations. The first of these lines should contain 2^k space-separated distinct integers p_{1}, p_{2}, …, p_{2^k}, and the second line should contain 2^k space-separated distinct integers q_{1}, q_{2}, …, q_{2^k}.

All elements of p and q should be between 0 and 2^k - 1, inclusive; p_i ⊕ q_i should be equal to a_i for all i such that 1 ≤ i ≤ 2^k. If there are several possible solutions, you can print any.

Examples

Input


2
0 1 2 3


Output


Shi
2 0 1 3 
2 1 3 0 


Input


2
0 0 0 0


Output


Shi
0 1 2 3 
0 1 2 3 


Input


2
0 1 2 2


Output


Fou

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N;
int A[1 << 12];
int P[1 << 12], RP[1 << 12];
int Q[1 << 12];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  N = 1 << N;
  x = 0;
  for (i = 0; i < (N); i++) {
    cin >> A[i];
    x ^= A[i];
    P[i] = Q[i] = RP[i] = i;
  }
  if (x) return (void)printf("Fou\n");
  for (i = 0; i < (N); i++) {
    x = i;
    while ((P[x] ^ Q[x]) != A[x]) {
      y = RP[A[x] ^ Q[x]];
      RP[P[x]] = y;
      RP[P[y]] = x;
      swap(P[y], P[x]);
      if (y > i) {
        break;
      }
      swap(Q[y], Q[i + 1]);
      x = y;
    }
  }
  cout << "Shi" << endl;
  for (i = 0; i < (N); i++) cout << P[i] << " ";
  cout << endl;
  for (i = 0; i < (N); i++) cout << Q[i] << " ";
  cout << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}