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 n, p[5000], q[5000], d[5000], bel[5000], a[5000];
void work(int x, int y) {
  int tmp = q[x] ^ d[x];
  if (bel[tmp] == x) return;
  if (bel[tmp] == y) {
    swap(p[x], p[y]), swap(bel[p[x]], bel[p[y]]);
    return;
  }
  int pos = bel[tmp];
  swap(bel[tmp], bel[p[x]]);
  swap(p[pos], p[x]);
  swap(q[pos], q[y]);
  work(pos, y);
}
int main() {
  scanf("%d", &n);
  n = 1 << n;
  int sum = 0;
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), p[i] = q[i] = i - 1, bel[i - 1] = i, sum ^= a[i],
                        d[i] = 0;
  if (sum) return puts("Fou"), 0;
  for (int i = 1; i < n; i++) {
    if (a[i] != d[i]) {
      d[i + 1] ^= (a[i] ^ d[i]);
      d[i] = a[i];
      work(i, i + 1);
    }
  }
  puts("Shi");
  for (int i = 1; i <= n; i++) printf("%d ", p[i]);
  puts("");
  for (int i = 1; i <= n; i++) printf("%d ", q[i]);
  puts("");
}