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 q[1000010], v[1000010], f[1000010], p[1000010], a[1000010];
void lily(int x, int y) {
  int c = v[x] ^ p[x];
  int k = f[c];
  if (k == x) return;
  if (k == y) {
    swap(q[x], q[y]);
    swap(f[q[x]], f[q[y]]);
    return;
  }
  swap(q[x], q[k]);
  swap(f[q[x]], f[q[k]]);
  swap(p[k], p[y]);
  lily(k, y);
  return;
}
int main() {
  int k, i;
  scanf("%d", &k);
  for (i = 0; i < (1 << k); i++) scanf("%d", &a[i]);
  int sum = 0;
  for (i = 0; i < (1 << k); i++) {
    sum ^= a[i];
    f[i] = i;
    q[i] = i;
    p[i] = i;
  }
  if (sum != 0) {
    printf("Fou\n");
    return 0;
  }
  printf("Shi\n");
  for (i = 0; i < (1 << k) - 1; i++) {
    int d = a[i] ^ v[i];
    if (d) {
      v[i] ^= d;
      v[i + 1] ^= d;
      lily(i, i + 1);
    }
  }
  for (i = 0; i < (1 << k); i++) printf("%d ", q[i]);
  puts("");
  for (i = 0; i < (1 << k); i++) printf("%d ", p[i]);
  puts("");
}