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;
const int N = 13;
int n, L, a[1 << N], b[1 << N], c[1 << N], d[1 << N], val[1 << N], sum;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  L = 1 << n;
  for (int i = 0; i < L; ++i)
    cin >> val[i], sum ^= val[i], a[i] = b[i] = c[i] = d[i] = i;
  if (sum) {
    cout << "Fou\n";
    return 0;
  }
  cout << "Shi\n";
  for (int i = 0; i < L; ++i)
    for (int j = i; (a[j] ^ b[j]) != val[j];) {
      int k = c[val[j] ^ b[j]];
      swap(a[j], a[k]);
      swap(c[a[j]], c[a[k]]);
      if (k > i) break;
      swap(b[k], b[i + 1]);
      swap(d[b[k]], d[b[i + 1]]);
      j = k;
    }
  for (int i = 0; i < L; ++i) cout << a[i] << " \n"[i == L - 1];
  for (int i = 0; i < L; ++i) cout << b[i] << " \n"[i == L - 1];
  return 0;
}