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>
int P[5005], Q[5005], arr[5005], goal[5005], idx[5005];
void update(int i, int j) {
  if (Q[i] == (P[i] ^ arr[i])) return;
  if (Q[i] == (P[i] ^ arr[j])) {
    std::swap(Q[i], Q[j]);
    std::swap(P[i], P[j]);
    std::swap(idx[P[i]], idx[P[j]]);
  } else if (Q[i] == (P[j] ^ arr[i])) {
    std::swap(P[i], P[j]);
    std::swap(idx[P[i]], idx[P[j]]);
  } else if (Q[i] == (P[j] ^ arr[j]))
    std::swap(Q[i], Q[j]);
  else {
    int t = idx[arr[i] ^ Q[i]];
    std::swap(Q[t], Q[j]);
    std::swap(P[t], P[i]);
    std::swap(idx[P[t]], idx[P[i]]);
    update(t, j);
  }
}
int main() {
  int k;
  scanf("%d", &k);
  int n = 1 << k;
  for (int i = 0; i < n; i++) scanf("%d", goal + i);
  for (int i = 0; i < n; i++) P[i] = Q[i] = idx[i] = i;
  for (int i = 0; i + 1 < n; i++) {
    int j = i + 1;
    arr[j] ^= arr[i] ^ goal[i];
    arr[i] = goal[i];
    update(i, j);
  }
  if (arr[n - 1] != goal[n - 1])
    puts("Fou");
  else {
    puts("Shi");
    for (int i = 0; i < n; i++) printf("%d%c", P[i], " \n"[i + 1 == n]);
    for (int i = 0; i < n; i++) printf("%d%c", Q[i], " \n"[i + 1 == n]);
  }
  return 0;
}