#include <bits/stdc++.h>
using namespace std;
int p[5010], w[5010], a[5010], q[5010], n, k, b[5010];
void adjust(int x, int y) {
  int P = w[b[x] ^ q[x]];
  if (P == x) return;
  if (P == y) {
    swap(p[x], p[y]), w[p[x]] = x, w[p[y]] = y;
    return;
  }
  swap(p[x], p[P]), w[p[x]] = x, w[p[P]] = P;
  swap(q[P], q[y]);
  adjust(P, y);
}
int main() {
  scanf("%d", &k);
  n = 1 << k;
  int X = 0;
  for (int i = 0; i < n; ++i)
    scanf("%d", a + i), X ^= a[i], w[i] = p[i] = q[i] = i;
  if (X == 0)
    printf("Shi\n");
  else {
    printf("Fou\n");
    return 0;
  }
  for (int i = 0; i < n - 1; ++i)
    if (b[i] ^ a[i]) {
      b[i + 1] ^= b[i] ^ a[i];
      b[i] = a[i];
      adjust(i, i + 1);
    }
  for (int i = 0; i < n; ++i) printf("%d ", p[i]);
  putchar('\n');
  for (int i = 0; i < n; ++i) printf("%d ", q[i]);
}

### 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>
using namespace std;
int p[5010], w[5010], a[5010], q[5010], n, k, b[5010];
void adjust(int x, int y) {
  int P = w[b[x] ^ q[x]];
  if (P == x) return;
  if (P == y) {
    swap(p[x], p[y]), w[p[x]] = x, w[p[y]] = y;
    return;
  }
  swap(p[x], p[P]), w[p[x]] = x, w[p[P]] = P;
  swap(q[P], q[y]);
  adjust(P, y);
}
int main() {
  scanf("%d", &k);
  n = 1 << k;
  int X = 0;
  for (int i = 0; i < n; ++i)
    scanf("%d", a + i), X ^= a[i], w[i] = p[i] = q[i] = i;
  if (X == 0)
    printf("Shi\n");
  else {
    printf("Fou\n");
    return 0;
  }
  for (int i = 0; i < n - 1; ++i)
    if (b[i] ^ a[i]) {
      b[i + 1] ^= b[i] ^ a[i];
      b[i] = a[i];
      adjust(i, i + 1);
    }
  for (int i = 0; i < n; ++i) printf("%d ", p[i]);
  putchar('\n');
  for (int i = 0; i < n; ++i) printf("%d ", q[i]);
}

```