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;
void solve() {
  int n;
  cin >> n;
  n = 1 << n;
  vector<int> a(n), p(n), q(n), pos(n), v(n);
  int sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum ^= a[i];
    p[i] = q[i] = pos[i] = i;
  }
  if (sum != 0) {
    cout << "Fou";
    return;
  }
  function<void(int, int)> fix = [&](int i, int j) {
    int k = pos[a[i] ^ q[i]];
    if (k == i) return;
    swap(p[i], p[k]);
    swap(pos[p[i]], pos[p[k]]);
    if (k != j) swap(q[k], q[j]), fix(k, j);
  };
  for (int i = 0; i < n; i++) {
    if (v[i] != a[i]) {
      int d = v[i] ^ a[i];
      v[i] ^= d;
      v[i + 1] ^= d;
      fix(i, i + 1);
    }
  }
  cout << "Shi\n";
  for (auto& x : p) {
    cout << x << ' ';
  }
  cout << '\n';
  for (auto& x : q) {
    cout << x << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}