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 read() {
  int s = 0, t = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * t;
}
int n, p[1 << 13], q[1 << 13], a[1 << 13], v[1 << 13], d, id[1 << 13];
void fix(int x, int y) {
  int del = q[x] ^ v[x], t = id[del];
  if (t == x) return;
  if (t == y) return swap(p[x], p[y]), swap(id[p[x]], id[p[y]]), void();
  swap(id[del], id[p[x]]);
  swap(p[t], p[x]);
  swap(q[t], q[y]);
  fix(t, y);
}
int main() {
  n = 1 << read();
  for (int i = 0; i < n; i++) p[i] = q[i] = id[i] = i, d ^= (a[i] = read());
  if (d) return puts("Fou"), 0;
  puts("Shi");
  for (int i = 0; i < n; i++) {
    int x = v[i] ^ a[i];
    v[i] ^= x, v[i + 1] ^= x;
    fix(i, i + 1);
  }
  for (int i = 0; i < n; i++) printf("%d ", p[i]);
  printf("\n");
  for (int i = 0; i < n; i++) printf("%d ", q[i]);
  printf("\n");
  return 0;
}