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 f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
const int MAXN = (1 << 12) + 1;
int N, p[MAXN], res, q[MAXN], v[MAXN], Ma[MAXN], A[MAXN];
void fix(int x, int y) {
  int d = q[x] ^ v[x], k = Ma[d];
  if (k == x) return;
  if (k == y) {
    swap(p[x], p[y]), swap(Ma[p[x]], Ma[p[y]]);
    return;
  }
  swap(p[k], p[x]), swap(q[k], q[y]), swap(Ma[p[x]], Ma[p[k]]);
  fix(k, y);
}
int main() {
  N = (1 << read());
  for (int i = 0; i < N; i++) A[i] = read(), res ^= A[i];
  if (res) {
    printf("Fou\n");
    return 0;
  }
  for (int i = 0; i < N; i++) p[i] = q[i] = Ma[i] = i;
  printf("Shi\n");
  for (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 (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");
}