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>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
  int n, k, i;
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> k;
  int sum = 0;
  n = 1 << k;
  vector<int> a(n), p(n), q(n), pos(n);
  for (i = 0; i < n; i++) {
    cin >> a[i];
    p[i] = q[i] = pos[i] = i;
    sum ^= a[i];
  }
  if (sum) {
    cout << "Fou\n";
    return 0;
  }
  for (i = 0; i < n; i++) {
    int cur = i;
    while ((p[cur] ^ q[cur]) != a[cur]) {
      int nxt = pos[a[cur] ^ q[cur]];
      swap(p[cur], p[nxt]), swap(pos[p[cur]], pos[p[nxt]]);
      if (nxt > i) {
        break;
      }
      swap(q[i + 1], q[nxt]);
      cur = nxt;
    }
  }
  cout << "Shi\n";
  for (auto it : p) {
    cout << it << " ";
  }
  cout << "\n";
  for (auto it : q) {
    cout << it << " ";
  }
  return 0;
}