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 main() {
  int k;
  scanf("%d", &k);
  int n = 1 << k;
  vector<int> b(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &b[i]);
  }
  int sum = 0;
  for (int i = 0; i < n; ++i) {
    sum ^= b[i];
  }
  if (sum != 0) {
    puts("Fou");
    return 0;
  }
  puts("Shi");
  vector<int> p(n), q(n), id(n), a(n);
  for (int i = 0; i < n; ++i) {
    p[i] = i;
    q[i] = i;
    id[i] = i;
  }
  function<void(int, int)> fix = [&](int i, int j) {
    if (q[i] == (p[i] ^ a[i])) {
      return;
    } else if (q[i] == (p[i] ^ a[j])) {
      swap(q[i], q[j]);
      swap(p[i], p[j]);
      swap(id[p[i]], id[p[j]]);
    } else if (q[i] == (p[j] ^ a[i])) {
      swap(p[i], p[j]);
      swap(id[p[i]], id[p[j]]);
    } else if (q[i] == (p[j] ^ a[j])) {
      swap(q[i], q[j]);
    } else {
      int t = id[a[i] ^ q[i]];
      swap(p[t], p[i]);
      swap(id[p[t]], id[p[i]]);
      swap(q[t], q[i]);
      swap(q[t], q[j]);
      swap(q[i], q[j]);
      fix(t, j);
    }
  };
  for (int i = 0; i < n - 1; ++i) {
    a[i + 1] ^= (a[i] ^ b[i]);
    a[i] = b[i];
    fix(i, i + 1);
  }
  for (int i = 0; i < n; ++i) {
    printf("%d ", p[i]);
  }
  puts("");
  for (int i = 0; i < n; ++i) {
    printf("%d ", q[i]);
  }
  puts("");
  return 0;
}