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 d = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  for (; isdigit(c); c = getchar()) d = d * 10 + c - '0';
  return d;
}
const int N = 5000;
int K, U;
int f[N], g[N];
int v[N], d[N], a[N];
inline bool chkFou() {
  int T(0);
  for (int i = (0), ed_i = (U - 1); i <= ed_i; ++i) T ^= a[i];
  return puts(T ? "Fou" : "Shi"), T;
}
inline void fix(int x, int y) {
  int p(g[x] ^ v[x]), t(d[p]);
  if (t == x) return;
  if (t == y) {
    (d[f[x]] ^= d[f[y]] ^= d[f[x]] ^= d[f[y]]), (f[x] ^= f[y] ^= f[x] ^= f[y]);
    return;
  }
  (d[p] ^= d[f[x]] ^= d[p] ^= d[f[x]]);
  (f[t] ^= f[x] ^= f[t] ^= f[x]), (g[t] ^= g[y] ^= g[t] ^= g[y]);
  fix(t, y);
}
int main() {
  K = read();
  U = 1 << K;
  for (int i = (0), ed_i = (U - 1); i <= ed_i; ++i) {
    a[i] = read();
    f[i] = g[i] = d[i] = i;
  }
  if (chkFou()) return 0;
  for (int i = (U - 1), ed_i = (1); i >= ed_i; --i) {
    if (v[i] != a[i]) {
      int x(v[i] ^ a[i]);
      v[i] ^= x, v[i - 1] ^= x;
      fix(i - 1, i);
    }
  }
  for (int i = (0), ed_i = (U - 1); i <= ed_i; ++i) printf("%d ", f[i]);
  puts("");
  for (int i = (0), ed_i = (U - 1); i <= ed_i; ++i) printf("%d ", g[i]);
  puts("");
  return 0;
}