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 maxn = 5005;
int k, p[maxn], q[maxn], a[maxn], b[maxn], pos[maxn], sum;
void dfs(int p1, int p2) {
  if (!(a[p1] ^ p[p1] ^ q[p1]) || !(a[p2] ^ p[p2] ^ q[p2])) return;
  if (pos[a[p1] ^ q[p1]] == p2) {
    swap(p[p1], p[p2]);
    swap(pos[p[p1]], pos[p[p2]]);
    return;
  }
  int t = pos[a[p1] ^ q[p1]];
  swap(p[t], p[p1]);
  swap(pos[p[t]], pos[p[p1]]);
  swap(q[p2], q[t]);
  dfs(t, p2);
}
int main() {
  cin >> k;
  for (int i = 0; i <= (1 << k) - 1; i++) cin >> b[i], sum ^= b[i];
  for (int i = 0; i <= (1 << k) - 1; i++) p[i] = q[i] = pos[i] = i;
  if (sum) {
    cout << "Fou";
    return 0;
  }
  cout << "Shi" << '\n';
  for (int i = 0; i <= (1 << k) - 2; i++) {
    a[i + 1] ^= (a[i] ^ b[i]);
    a[i] = b[i];
    dfs(i, i + 1);
  }
  for (int i = 0; i <= (1 << k) - 1; i++) cout << p[i] << ' ';
  cout << '\n';
  for (int i = 0; i <= (1 << k) - 1; i++) cout << q[i] << ' ';
  return 0;
}