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 maxn = 12, maxm = 1 << maxn;
int n, m, a[maxm], x, y, p[maxm], q[maxm], r[maxm], s[maxm];
int main() {
  cin.tie(0), cout.tie(0), ios::sync_with_stdio(0);
  cin >> n, m = 1 << n;
  for (int i = 0; i < m; i++) cin >> a[i], x ^= a[i], p[i] = q[i] = r[i] = i;
  if (x) return cout << "Fou" << endl, 0;
  for (int i = 0; i < m; i++)
    if (s[i] != a[i]) {
      s[i + 1] ^= s[i] ^ a[i], x = i;
      while ((y = r[q[x] ^ a[x]]) != x && y != i + 1)
        swap(r[p[x]], r[p[y]]), swap(p[x], p[y]), swap(x, y),
            swap(q[x], q[i + 1]);
      if (y == i + 1) swap(r[p[x]], r[p[y]]), swap(p[x], p[y]);
    }
  cout << "Shi" << endl;
  for (int i = 0; i < m; i++) cout << p[i] << ' ';
  cout << endl;
  for (int i = 0; i < m; i++) cout << q[i] << ' ';
  cout << endl;
  return 0;
}