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>
const int N = 4100;
int n, l, pos;
int a[N], p[N], q[N], ip[N];
inline void swapP(int u, int v) {
  std::swap(ip[p[u]], ip[p[v]]), std::swap(p[u], p[v]);
}
void adjust(int i) {
  if (!(p[i] ^ q[i] ^ a[i])) return;
  int j = ip[q[i] ^ a[i]];
  if (swapP(i, j), j < pos) std::swap(q[j], q[pos]), adjust(j);
}
int main() {
  int i;
  scanf("%d", &l), n = 1 << l;
  for (i = 0; i < n; ++i) scanf("%d", a + i), p[i] = q[i] = ip[i] = i;
  if (std::accumulate(a, a + n, 0, std::bit_xor<int>())) return puts("Fou"), 0;
  for (pos = 1; pos < n; ++pos) adjust(pos - 1);
  puts("Shi");
  for (i = 0; i < n; ++i) printf("%d%c", p[i], i == n - 1 ? 10 : 32);
  for (i = 0; i < n; ++i) printf("%d%c", q[i], i == n - 1 ? 10 : 32);
  return 0;
}