#include <bits/stdc++.h>
int main() {
  static int aa[4096], pp[4096], qq[4096], ii[4096];
  int n, k, i, u, v, tmp, a;
  scanf("%d", &k);
  n = 1 << k;
  a = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &aa[i]);
    a ^= aa[i];
  }
  if (a) {
    printf("Fou\n");
    return 0;
  }
  for (i = 0; i < n; i++) pp[i] = qq[i] = ii[i] = i;
  for (i = 0; i < n - 1; i++)
    for (u = i; u <= i && aa[u] ^ pp[u] ^ qq[u]; u = v) {
      v = ii[aa[u] ^ pp[u]];
      tmp = qq[u], qq[u] = qq[v], qq[v] = tmp;
      tmp = ii[qq[u]], ii[qq[u]] = ii[qq[v]], ii[qq[v]] = tmp;
      tmp = pp[v], pp[v] = pp[n - 1], pp[n - 1] = tmp;
    }
  printf("Shi\n");
  for (i = 0; i < n; i++) printf("%d ", pp[i]);
  printf("\n");
  for (i = 0; i < n; i++) printf("%d ", qq[i]);
  printf("\n");
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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>
int main() {
  static int aa[4096], pp[4096], qq[4096], ii[4096];
  int n, k, i, u, v, tmp, a;
  scanf("%d", &k);
  n = 1 << k;
  a = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &aa[i]);
    a ^= aa[i];
  }
  if (a) {
    printf("Fou\n");
    return 0;
  }
  for (i = 0; i < n; i++) pp[i] = qq[i] = ii[i] = i;
  for (i = 0; i < n - 1; i++)
    for (u = i; u <= i && aa[u] ^ pp[u] ^ qq[u]; u = v) {
      v = ii[aa[u] ^ pp[u]];
      tmp = qq[u], qq[u] = qq[v], qq[v] = tmp;
      tmp = ii[qq[u]], ii[qq[u]] = ii[qq[v]], ii[qq[v]] = tmp;
      tmp = pp[v], pp[v] = pp[n - 1], pp[n - 1] = tmp;
    }
  printf("Shi\n");
  for (i = 0; i < n; i++) printf("%d ", pp[i]);
  printf("\n");
  for (i = 0; i < n; i++) printf("%d ", qq[i]);
  printf("\n");
  return 0;
}

```