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;
const int N = 5005;
int n, aa[N], bb[N], a[N], v[N], bel[N], d;
void fix(int x, int y) {
  int p = bb[x] ^ v[x], t = bel[p];
  if (t == x) return;
  if (t == y) {
    swap(aa[x], aa[y]);
    swap(bel[aa[x]], bel[aa[y]]);
    return;
  }
  swap(bel[p], bel[aa[x]]);
  swap(aa[t], aa[x]);
  swap(bb[t], bb[y]);
  fix(t, y);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  n = 1 << n;
  for (int i = 0; i < n; i++)
    cin >> a[i], d ^= a[i], aa[i] = bb[i] = bel[i] = i;
  if (d) {
    cout << "Fou" << endl;
    return 0;
  }
  cout << "Shi" << endl;
  for (int i = 0; i + 1 < n; i++)
    if (v[i] != a[i]) {
      int x = v[i] ^ a[i];
      v[i] ^= x;
      v[i + 1] ^= x;
      fix(i, i + 1);
    }
  for (int i = 0; i < n; i++) cout << aa[i] << ' ';
  cout << endl;
  for (int i = 0; i < n; i++) cout << bb[i] << ' ';
  return 0;
}