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;
using cat = long long;
void sub(int pos, vector<int>& occ, vector<int>& occT, int& dif) {
  dif -= (((occ[pos] - occT[pos]) < 0) ? -(occ[pos] - occT[pos])
                                       : (occ[pos] - occT[pos]));
  occ[pos]--;
  dif += (((occ[pos] - occT[pos]) < 0) ? -(occ[pos] - occT[pos])
                                       : (occ[pos] - occT[pos]));
}
void add(int pos, vector<int>& occ, vector<int>& occT, int& dif) {
  dif -= (((occ[pos] - occT[pos]) < 0) ? -(occ[pos] - occT[pos])
                                       : (occ[pos] - occT[pos]));
  occ[pos]++;
  dif += (((occ[pos] - occT[pos]) < 0) ? -(occ[pos] - occT[pos])
                                       : (occ[pos] - occT[pos]));
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  unsigned K;
  cin >> K;
  int N = 1 << K;
  vector<int> A(N), occT(N, 0);
  int x = 0;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    x ^= A[i];
    occT[A[i]]++;
  }
  srand(std::chrono::steady_clock::now().time_since_epoch().count());
  if (x != 0) {
    cout << "Fou\n";
    return 0;
  }
  vector<int> P(N), occ(N, 1);
  for (int i = 0; i < (1 << (K - 1)); i++) {
    P[2 * i] = i;
    P[N - 1 - 2 * i] = (1 << (K - 1)) + i;
  }
  if (K % 2 != 0)
    for (int i = 0; i < (1 << (K - 2)); i++)
      swap(P[N / 2 + 2 * i], P[N / 2 + 2 * i + 1]);
  int dif = 0;
  for (int i = 0; i < N; i++) {
    dif +=
        (((occT[i] - occ[i]) < 0) ? -(occT[i] - occ[i]) : (occT[i] - occ[i]));
  }
  int step;
  for (step = 0;; step++) {
    if (dif == 0) break;
    vector<int> p(N);
    for (int i = 0; i < N; i++) {
      p[i] = i;
      swap(p[i], p[rand() % (i + 1)]);
    }
    bool fail = true;
    for (int k = 0; k < 2; k++)
      if (k == 0 || fail)
        for (int i = 0; i < N; i++)
          if (occ[p[i] ^ P[p[i]]] > occT[p[i] ^ P[p[i]]])
            for (int j = 0; j < i; j++) {
              int dif_nw = dif;
              sub(p[i] ^ P[p[i]], occ, occT, dif_nw);
              sub(p[j] ^ P[p[j]], occ, occT, dif_nw);
              add(p[i] ^ P[p[j]], occ, occT, dif_nw);
              add(p[j] ^ P[p[i]], occ, occT, dif_nw);
              if (dif_nw < dif || (dif_nw <= dif && fail)) {
                swap(P[p[i]], P[p[j]]);
                dif = dif_nw;
                if (k == 0) fail = false;
                continue;
              }
              occ[p[i] ^ P[p[i]]]++;
              occ[p[j] ^ P[p[j]]]++;
              occ[p[i] ^ P[p[j]]]--;
              occ[p[j] ^ P[p[i]]]--;
            }
  }
  cout << "Shi\n";
  vector<pair<int, int> > ans(N);
  for (int i = 0; i < N; i++) ans[i] = {i, P[i]};
  for (int i = 0; i < N; i++)
    for (int j = i; j < N; j++)
      if ((ans[j].first ^ ans[j].second) == A[i]) {
        swap(ans[i], ans[j]);
        break;
      }
  for (int i = 0; i < N; i++)
    cout << ans[i].first << ((i == N - 1) ? "\n" : " ");
  for (int i = 0; i < N; i++)
    cout << ans[i].second << ((i == N - 1) ? "\n" : " ");
  return 0;
}