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 = (1 << 12) + 5;
int n, a[maxn], b[2][maxn], pos[2][maxn];
mt19937 rnd(time(0));
inline int get(int opt) {
  int res = rnd() % n;
  while (pos[opt][res]) res = rnd() % n;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  n = 1 << n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int sum = 0;
  for (int i = 1; i <= n; i++) sum ^= a[i];
  if (sum) return puts("Fou"), 0;
  cout << "Shi" << endl;
  for (int i = 1; i <= n; i++) {
    b[0][i] = get(0);
    pos[0][b[0][i]] = i;
    b[1][i] = a[i] ^ b[0][i];
    int npos = i, opt = 1;
    while (1) {
      int now = b[opt][npos];
      if (!pos[opt][now]) {
        pos[opt][now] = npos;
        break;
      } else {
        int pre = pos[opt][now];
        pos[opt][now] = npos;
        npos = pre, b[opt][npos] = get(opt);
        pos[opt][b[opt][npos]] = npos;
        pos[opt ^ 1][b[opt ^ 1][npos]] = 0;
        b[opt ^ 1][npos] = a[npos] ^ b[opt][npos];
        opt ^= 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << b[0][i] << " ";
  cout << endl;
  for (int i = 1; i <= n; i++) cout << b[1][i] << " ";
  cout << endl;
  return 0;
}