#include <bits/stdc++.h>
using std::swap;
int n, a[10010], b[10010], c[10010], pa[10010], pb[10010], now;
int main() {
  scanf("%d", &n);
  n = 1 << n;
  for (int i = 0; i < n; i++)
    scanf("%d", &c[i]), a[i] = b[i] = pa[i] = pb[i] = i, now ^= c[i];
  if (now != 0) return puts("Fou"), 0;
  puts("Shi");
  for (int i = 0; i < n; i++)
    for (int j = i; (a[j] ^ b[j]) != c[j];) {
      int x = pa[c[j] ^ b[j]];
      swap(a[j], a[x]);
      swap(pa[a[j]], pa[a[x]]);
      if (x > i) break;
      swap(b[x], b[i + 1]);
      swap(pb[b[x]], pb[b[i + 1]]);
      j = x;
    }
  for (int i = 0; i < n; i++) printf("%d ", a[i]);
  puts("");
  for (int i = 0; i < n; i++) printf("%d ", b[i]);
}

### Prompt

Create a solution in Cpp for the following 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

### Response

```cpp
#include <bits/stdc++.h>
using std::swap;
int n, a[10010], b[10010], c[10010], pa[10010], pb[10010], now;
int main() {
  scanf("%d", &n);
  n = 1 << n;
  for (int i = 0; i < n; i++)
    scanf("%d", &c[i]), a[i] = b[i] = pa[i] = pb[i] = i, now ^= c[i];
  if (now != 0) return puts("Fou"), 0;
  puts("Shi");
  for (int i = 0; i < n; i++)
    for (int j = i; (a[j] ^ b[j]) != c[j];) {
      int x = pa[c[j] ^ b[j]];
      swap(a[j], a[x]);
      swap(pa[a[j]], pa[a[x]]);
      if (x > i) break;
      swap(b[x], b[i + 1]);
      swap(pb[b[x]], pb[b[i + 1]]);
      j = x;
    }
  for (int i = 0; i < n; i++) printf("%d ", a[i]);
  puts("");
  for (int i = 0; i < n; i++) printf("%d ", b[i]);
}

```