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 get() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-') {
    int s = 0;
    while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
    return s;
  }
  int s = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
  return s;
}
const int N = (1 << 12) + 5;
int n;
int p[N], q[N];
int a[N], v[N];
int pos[N];
void solve(int i, int j) {
  if (a[i] == (p[i] ^ q[i]) && a[j] == (p[j] ^ a[j])) return;
  int t = pos[a[i] ^ q[i]];
  swap(p[i], p[t]);
  pos[p[i]] = i, pos[p[j]] = j, pos[p[t]] = t;
  swap(q[t], q[j]);
  solve(t, j);
}
int main() {
  int k = get();
  n = (1 << k);
  int sum = 0;
  for (int i = 0; i <= n - 1; i++) {
    v[i] = get();
    sum ^= v[i];
  }
  if (sum != 0) return printf("Fou\n"), 0;
  for (int i = 0; i <= n - 1; i++) p[i] = q[i] = pos[i] = i, a[i] = 0;
  for (int i = 0; i <= n - 2; i++) {
    int x = a[i] ^ v[i];
    a[i] ^= x, a[i + 1] ^= x;
    solve(i, i + 1);
  }
  printf("Shi\n");
  for (int i = 0; i <= n - 1; i++) printf("%d ", p[i]);
  putchar('\n');
  for (int i = 0; i <= n - 1; i++) printf("%d ", q[i]);
  putchar('\n');
  return 0;
}