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 main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int k;
  cin >> k;
  vector<int> v(1 << k);
  int xors = 0;
  for (int i = 0; i < (1 << k); ++i) {
    cin >> v[i];
    xors ^= v[i];
  }
  if (xors == 0) {
    cout << "Shi\n";
    vector<int> a(1 << k), b(1 << k), pos(1 << k);
    iota(a.begin(), a.end(), 0);
    iota(b.begin(), b.end(), 0);
    iota(pos.begin(), pos.end(), 0);
    for (int i = 0; i < (1 << k); ++i) {
      int x = i;
      while (a[x] ^ b[x] ^ v[x]) {
        int y = pos[v[x] ^ b[x]];
        swap(a[x], a[y]);
        swap(pos[a[x]], pos[a[y]]);
        swap(b[i + 1], b[y]);
        x = y;
      }
    }
    for (int i : a) {
      cout << i << ' ';
    }
    cout << '\n';
    for (int i : b) {
      cout << i << ' ';
    }
    cout << '\n';
  } else {
    cout << "Fou\n";
  }
}