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;
int a[maxn];
int b[maxn];
int p[maxn], q[maxn];
int rp[maxn], rq[maxn];
int k;
int main() {
  scanf("%d", &k);
  int n = 1 << k;
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sum ^= a[i];
  }
  if (sum) {
    printf("Fou");
    return 0;
  }
  printf("Shi\n");
  for (int i = 1; i <= n; i++) {
    p[i] = q[i] = i - 1;
    rp[i - 1] = rq[i - 1] = i;
  }
  for (int i = 1; i < n; i++) {
    int indi = i;
    int indj = i + 1;
    b[indj] ^= a[indi] ^ b[indi];
    b[indi] = a[indi];
    while (indi != indj) {
      int t = rp[b[indi] ^ q[indi]];
      rp[p[t]] = indi;
      rq[q[t]] = indj;
      rp[p[indi]] = t;
      rq[q[indj]] = t;
      swap(p[t], p[indi]);
      swap(q[t], q[indj]);
      indi = t;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", p[i]);
  }
  printf("\n");
  for (int i = 1; i <= n; i++) {
    printf("%d ", q[i]);
  }
  return 0;
}