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 void read(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void read(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void readc(char &x) {
  char c;
  while (((c = getchar()) == ' ') || c == '\n')
    ;
  x = c;
}
int n, m, i, j, a[1 << 12], b[1 << 12], pa[1 << 12], push_back[1 << 12],
    c[1 << 12], sum;
int main() {
  read(n);
  for (((i)) = (0); ((i)) <= (((int)((1 << n))) - 1); ((i))++) {
    a[i] = b[i] = pa[i] = push_back[i] = i;
    read(c[i]);
    sum ^= c[i];
  }
  if (sum) {
    puts("Fou");
    return 0;
  }
  for (((i)) = (0); ((i)) <= (((int)((1 << n))) - 1); ((i))++) {
    int t = i;
    while ((a[t] ^ b[t]) != c[t]) {
      int p = pa[b[t] ^ c[t]];
      swap(a[t], a[p]);
      swap(pa[a[t]], pa[a[p]]);
      if (p > i) break;
      swap(b[p], b[i + 1]);
      swap(push_back[b[p]], push_back[b[i + 1]]);
      t = p;
    }
  }
  puts("Shi");
  for (((i)) = (0); ((i)) <= (((int)((1 << n))) - 1); ((i))++)
    printf("%d ", a[i]);
  puts("");
  for (((i)) = (0); ((i)) <= (((int)((1 << n))) - 1); ((i))++)
    printf("%d ", b[i]);
  puts("");
  return 0;
}