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 = 1 << 12;
int n, s, len, x[N], p[N], q[N], atp[N], atq[N];
mt19937 rng(20040725 ^ time(0));
inline int gen(int *at) {
  static int x;
  do x = rng() % len;
  while (~at[x]);
  return x;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n, len = 1 << n;
  for (int i = 0; i < len; i++) cin >> x[i], s ^= x[i];
  if (s) return cout << "Fou\n", 0;
  while (true) {
    int cnt = 0;
    memset(atp, -1, len << 2), memset(atq, -1, len << 2);
    for (int u, i = 0; i < len; i++) {
      p[i] = -1;
      for (u = i; ~u; swap(u, atq[q[u]])) {
        if (~p[u]) atp[p[u]] = -1;
        p[u] = gen(atp), atp[p[u]] = u, q[u] = x[u] ^ p[u];
        if (++cnt == 10000000) goto nxt;
      }
    }
    break;
  nxt:
    continue;
  }
  cout << "Shi\n";
  for (int i = 0; i < len; i++) cout << p[i] << " \n"[i + 1 == len];
  for (int i = 0; i < len; i++) cout << q[i] << " \n"[i + 1 == len];
}