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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 100010;
const int INF = 2147483600;
int K;
int p[MAXN + 1], q[MAXN + 1], pos[MAXN + 1];
int N, a[MAXN + 1];
inline void Fix(int i, int j, int x) {
  int k = pos[x ^ p[i]];
  if (k == i) return;
  swap(p[i], p[k]);
  swap(pos[p[i]], pos[p[k]]);
  if (k == j) return;
  swap(q[j], q[k]);
  Fix(k, j, x ^ q[j] ^ q[k]);
  return;
}
int main() {
  K = read();
  N = (1 << K);
  int s = 0;
  for (int i = 0; i < N; i++)
    a[i] = read(), p[i] = i, q[i] = i, pos[i] = i, s ^= a[i];
  if (s) {
    puts("Fou");
    return 0;
  }
  for (int i = 0; i < N - 1; i++) {
    Fix(i, i + 1, p[i] ^ q[i] ^ a[i]);
  }
  puts("Shi");
  for (int i = 0; i < N; i++) printf("%d ", p[i]);
  printf("\n");
  for (int i = 0; i < N; i++) printf("%d ", q[i]);
  return 0;
}