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;
const int N = 5010;
int a[N], b[N], p[N], q[N];
int idx[N];
inline void fix(int i, int j) {
  if (a[i] == (p[i] ^ q[i]))
    return;
  else if (q[i] == (p[i] ^ a[j])) {
    swap(q[i], q[j]);
    swap(p[i], p[j]);
    swap(idx[p[i]], idx[p[j]]);
  } else if (q[i] == (p[j] ^ a[i])) {
    swap(p[i], p[j]);
    swap(idx[p[i]], idx[p[j]]);
  } else if (q[i] == (p[j] ^ a[j])) {
    swap(q[i], q[j]);
  } else {
    int t = idx[a[i] ^ q[i]];
    swap(idx[p[t]], idx[p[i]]);
    swap(p[t], p[i]);
    swap(q[t], q[i]);
    swap(q[t], q[j]);
    swap(q[i], q[j]);
    fix(t, j);
  }
}
int main() {
  cin >> n;
  n = 1 << n;
  int tot = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &b[i]);
    idx[i] = p[i] = q[i] = i;
    tot ^= b[i];
  }
  if (tot) {
    puts("Fou");
    return 0;
  } else {
    puts("Shi");
    for (int i = 0; i + 1 < n; 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("");
  }
}