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;
const int N = 1 << 12 | 1;
int n, a[N], b[N], c[N], d[N], to[N];
inline void fix(int x, int y) {
  int val = d[x] ^ c[x], k = to[val];
  if (k == x) return;
  if (k == y) {
    swap(b[x], b[y]);
    swap(to[b[x]], to[b[y]]);
    return;
  }
  swap(b[x], b[k]);
  swap(c[k], c[y]);
  swap(to[b[x]], to[b[k]]);
  fix(k, y);
}
int main() {
  scanf("%d", &n);
  n = 1 << n;
  int sum = 0;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), sum ^= a[i];
  if (sum != 0) {
    puts("Fou");
    return 0;
  }
  puts("Shi");
  for (int i = 0; i < n; i++) b[i] = c[i] = i, d[i] = 0, to[i] = i;
  for (int i = 0; i < n; i++)
    if (d[i] != a[i]) {
      d[i + 1] ^= d[i] ^ a[i];
      d[i] = a[i];
      fix(i, i + 1);
    }
  for (int i = 0; i < n; i++) printf("%d ", b[i]);
  puts("");
  for (int i = 0; i < n; i++) printf("%d ", c[i]);
}