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[1 << 12], p[1 << 12], q[1 << 12], posp[1 << 12], posq[1 << 12];
int main() {
  int i;
  int k, x = 0;
  scanf("%d", &k);
  for (i = 0; i < (1 << k); i++) scanf("%d", &a[i]), x ^= a[i];
  if (x != 0) {
    printf("Fou\n");
    return 0;
  }
  for (i = 0; i < (1 << k); i++) p[i] = q[i] = posp[i] = posq[i] = i;
  for (i = 0; i < (1 << k) - 1; i++) {
    while ((a[i] ^ p[i] ^ q[i]) != 0) {
      x = a[i] ^ p[i] ^ q[i];
      while (1) {
        int u = i, c = 0;
        do {
          c++;
          if (c & 1)
            u = posp[p[u] ^ x];
          else
            u = posq[q[u] ^ x];
        } while ((u != i) && (u != i + 1));
        if (u == i + 1) {
          u = i, c = 0;
          do {
            c++;
            if (c & 1) {
              int v = posp[p[u] ^ x];
              swap(p[u], p[v]), swap(posp[p[u]], posp[p[v]]);
              u = v;
            } else {
              int v = posq[q[u] ^ x];
              swap(q[u], q[v]), swap(posq[q[u]], posq[q[v]]);
              u = v;
            }
          } while (u != i + 1);
          break;
        } else
          x = rand() % (1 << k);
      }
    }
  }
  printf("Shi\n");
  for (i = 0; i < (1 << k); i++) printf("%d ", p[i]);
  printf("\n");
  for (i = 0; i < (1 << k); i++)
    printf("%d ", q[i]), assert((p[i] ^ q[i] ^ a[i]) == 0);
  printf("\n");
  return 0;
}