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 = 5010;
int n, a[N], t[N], p[N], q[N], posp[N], sum;
void solve(int i, int j) {
  int k = posp[t[i] ^ q[i]];
  if (k == i) {
    return;
  }
  if (k == j) {
    swap(posp[p[i]], posp[p[j]]), swap(p[i], p[j]);
    return;
  }
  swap(posp[p[i]], posp[p[k]]), swap(p[i], p[k]), swap(q[j], q[k]);
  solve(k, j);
}
int main() {
  scanf("%d", &n);
  n = 1 << n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    p[i] = q[i] = posp[i] = i;
    sum ^= a[i];
  }
  if (sum) {
    puts("Fou");
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] != t[i]) {
      int dt = a[i] ^ t[i];
      t[i] ^= dt, t[i + 1] ^= dt;
      solve(i, 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 ", q[i]);
  }
  puts("");
}