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 std::cin;
using std::cout;
const int maxn = 200200;
int a[maxn], b[maxn], c[maxn], ib[maxn];
int k;
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> k;
  int sum = 0;
  for (int i = 0; i < 1 << k; ++i) {
    cin >> a[i];
    sum ^= a[i];
    ib[i] = b[i] = c[i] = i;
  }
  if (sum) {
    cout << "Fou" << '\n';
    return 0;
  }
  for (int i = 0; i < 1 << k; ++i) {
    int x = i;
    for (; a[x] ^ b[x] ^ c[x];) {
      int y = ib[a[x] ^ c[x]];
      std::swap(b[x], b[y]);
      std::swap(ib[b[x]], ib[b[y]]);
      if (y > i) {
        break;
      }
      std::swap(c[y], c[i + 1]);
      x = y;
    }
  }
  cout << "Shi" << '\n';
  for (int i = 0; i < 1 << k; ++i) {
    cout << b[i] << " \n"[i + 1 == 1 << k];
  }
  for (int i = 0; i < 1 << k; ++i) {
    cout << c[i] << " \n"[i + 1 == 1 << k];
  }
}