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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count() *
               ((uint64_t) new char | 1));
int K, N;
vector<int> A, B, Z;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> K;
  N = 1 << K;
  Z.resize(N);
  int all_xor = 0;
  for (auto &z : Z) {
    cin >> z;
    all_xor ^= z;
  }
  if (all_xor != 0) {
    cout << "Fou" << '\n';
    return 0;
  }
  A.resize(N);
  B.resize(N);
  vector<int> location(N);
  for (int i = 0; i < N; i++) A[i] = B[i] = i;
  shuffle(A.begin(), A.end(), rng);
  shuffle(B.begin(), B.end(), rng);
  for (int i = 0; i < N; i++) location[B[i]] = i;
  for (int i = 0; i < N; i++) {
    int u = i;
    while (Z[u] != (A[u] ^ B[u])) {
      int v = location[A[u] ^ Z[u]];
      swap(B[u], B[v]);
      swap(location[B[u]], location[B[v]]);
      if (v > i) {
        break;
      } else {
        swap(A[v], A[i + 1]);
        u = v;
      }
    }
  }
  cout << "Shi" << '\n';
  for (int i = 0; i < N; i++) cout << A[i] << (i < N - 1 ? ' ' : '\n');
  for (int i = 0; i < N; i++) cout << B[i] << (i < N - 1 ? ' ' : '\n');
  return 0;
}