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 N = 1 << 12;
template <typename T>
inline void read(T &x) {
  int ch = getchar();
  x = 0;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
}
int n, k, sum, a[N], p[N], q[N], pre[N], v[N];
inline void fix(int i, int j) {
  int d = q[i] ^ v[i], k = pre[d];
  if (k == i) return;
  if (k == j) return swap(p[i], p[j]), swap(pre[p[i]], pre[p[j]]);
  swap(p[k], p[i]);
  swap(q[k], q[j]);
  swap(pre[p[i]], pre[p[k]]);
  fix(k, j);
}
int main() {
  read(k);
  n = 1 << k;
  for (register int i = 0; i < n; ++i)
    read(a[i]), p[i] = q[i] = pre[i] = i, sum ^= a[i];
  if (sum) return puts("Fou"), 0;
  puts("Shi");
  for (register int i = 0; i < n - 1; ++i)
    if (v[i] != a[i]) {
      int d = v[i] ^ a[i];
      v[i] ^= d;
      v[i + 1] ^= d;
      fix(i, i + 1);
    }
  for (register int i = 0; i < n; ++i) printf("%d%c", p[i], " \n"[i == n - 1]);
  for (register int i = 0; i < n; ++i) printf("%d%c", q[i], " \n"[i == n - 1]);
}