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 = 1e5 + 5;
int n, k, s, a[MAXN], v[MAXN], p[MAXN], q[MAXN], pos[MAXN];
inline int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
void fix(int i, int j) {
  int t = pos[v[i] ^ q[i]];
  if (t == i) return;
  if (t == j) {
    swap(p[i], p[j]), v[i] = p[i] ^ q[i];
    swap(pos[p[i]], pos[p[j]]);
    return v[j] = p[j] ^ q[j], void();
  }
  swap(p[t], p[i]), swap(q[t], q[j]);
  swap(pos[p[t]], pos[p[i]]), fix(t, j);
}
int main() {
  k = read(), n = 1 << k;
  for (int i = 0; i < n; i++) s ^= (a[i] = read());
  if (s) puts("Fou"), exit(0);
  puts("Shi");
  for (int i = 0; i < n; i++) p[i] = q[i] = pos[i] = i;
  for (int i = 0; i < n - 1; i++) {
    v[n] = a[i] ^ v[i], v[i] ^= v[n], v[i + 1] ^= v[n];
    fix(i, i + 1);
  }
  for (int i = 0; i < n; i++) printf("%d%c", p[i], " \n"[i == n - 1]);
  for (int i = 0; i < n; i++) printf("%d%c", q[i], " \n"[i == n - 1]);
  return 0;
}