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;
int a[N], p[N], f1[N], f2[N];
int main() {
  int k;
  scanf("%d", &k);
  int n = 1 << k, sum = n, X = 0;
  for (int i = 0; i < n; i++) scanf("%d", a + i), X ^= a[i];
  if (X) puts("Fou"), exit(0);
  memset(p, -1, sizeof(p));
  while (sum) {
    int t = rand() % sum + 1;
    int x = 0, c1 = 0, I = 0;
    for (int i = 0; i < n; i++)
      if (!f1[i]) {
        c1++;
        if (c1 == t) x = i;
      }
    f1[x] = 1;
    c1 = 0;
    for (int i = 0; i < n; i++)
      if (!~p[i] && !f2[a[i] ^ x]) c1++;
    if (c1) {
      t = rand() % c1 + 1;
      c1 = 0;
      for (int i = 0; i < n; i++)
        if (!~p[i] && !f2[a[i] ^ x]) {
          c1++;
          if (c1 == t) I = i;
        }
      p[I] = x;
      f2[a[I] ^ p[I]] = 1;
      sum--;
    } else {
      c1 = 0;
      for (int i = 0; i < n; i++)
        if (!~p[i]) c1++;
      t = rand() % c1 + 1;
      c1 = 0;
      for (int i = 0; i < n; i++)
        if (!~p[i]) {
          c1++;
          if (c1 == t) I = i;
        }
      for (int i = 0; i < n; i++)
        if (~p[i] && (p[i] ^ a[i]) == (x ^ a[I])) {
          f1[p[i]] = f2[p[i] ^ a[i]] = 0;
          p[i] = -1;
        }
      p[I] = x;
      f1[p[I]] = f2[a[I] ^ p[I]] = 1;
    }
  }
  puts("Shi");
  for (int i = 0; i < n; i++) printf("%d ", p[i]);
  puts("");
  for (int i = 0; i < n; i++) printf("%d ", p[i] ^ a[i]);
  puts("");
  return 0;
}