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;
int a[10010], p[10010], q[10010], need[10010], pos_q[10010];
void solve(int x, int y) {
  if (a[x] == (p[x] ^ q[x]) && a[y] == (p[y] ^ q[y])) return;
  int tmp = pos_q[a[x] ^ p[x]];
  swap(q[x], q[tmp]);
  swap(p[y], p[tmp]);
  pos_q[q[x]] = x;
  pos_q[q[tmp]] = tmp;
  solve(tmp, y);
}
int main() {
  int k, n, res = 0;
  scanf("%d", &k);
  n = (1 << k) - 1;
  for (int i = 0; i <= n; i++) {
    scanf("%d", &need[i]);
    res ^= need[i];
  }
  if (res) {
    puts("Fou");
    return 0;
  }
  for (int i = 0; i <= n; i++) {
    p[i] = q[i] = i;
    pos_q[i] = i;
  }
  for (int i = 0; i <= n - 1; i++) {
    int tmp = a[i] ^ need[i];
    a[i] ^= tmp;
    a[i + 1] ^= tmp;
    solve(i, i + 1);
  }
  puts("Shi");
  for (int i = 0; i <= n; i++) printf("%d ", p[i]);
  puts("");
  for (int i = 0; i <= n; i++) printf("%d ", q[i]);
  puts("");
  return 0;
}