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 k, A[1 << 12], B[1 << 12], Sym[1 << 12][2], Ans[1 << 12][2];
void Do(int a, int b, int s) {
  swap(Ans[a][s], Ans[b][s]), swap(Sym[Ans[a][s]][s], Sym[Ans[b][s]][s]);
}
int main() {
  scanf("%d", &k);
  for (int i = 0; i < (1 << k); i++)
    scanf("%d", &A[i]), Ans[i][0] = (Ans[i][1] = (Sym[i][0] = (Sym[i][1] = i)));
  int sum = 0;
  for (int i = 0; i < (1 << k); i++) sum ^= A[i];
  if (sum) {
    puts("Fou");
    return 0;
  }
  for (int i = 1; i < (1 << k); i++)
    if (A[i]) {
      B[0] ^= A[i], B[i] = A[i];
      int a = 0, b = i;
      while (Ans[a][0] ^ Ans[a][1] ^ B[a]) {
        int s = Sym[Ans[a][0] ^ B[a]][1];
        if (s == b) {
          Do(a, b, 1);
          break;
        }
        Do(a, s, 1), Do(b, s, 0);
        a = s;
      }
    }
  puts("Shi");
  for (int i = 0; i < (1 << k); i++) printf("%d ", Ans[i][0]);
  putchar('\n');
  for (int i = 0; i < (1 << k); i++) printf("%d ", Ans[i][1]);
  putchar('\n');
  return 0;
}