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;
int fl[4130];
int a[4130];
int p[4130], q[4130], pl[4130];
int nu[4130], ncnt;
int nr[4130], ucnt;
int main() {
  int k;
  cin >> k;
  int n = (1 << k);
  int ns = 0;
  for (int i = 0; i < n; i++) cin >> a[i], ns ^= a[i];
  memset(p, -1, sizeof(p));
  memset(q, -1, sizeof(q));
  if (ns) {
    cout << "Fou" << endl;
    return 0;
  }
  while (1) {
    memset(pl, -1, sizeof(pl));
    memset(fl, 0, sizeof(fl));
    for (int i = 0; i < n; i++)
      if (p[i] != -1) fl[p[i]] = 1, pl[q[i]] = i;
    ncnt = 0, ucnt = 0;
    for (int i = 0; i < n; i++)
      if (!fl[i]) nu[ncnt++] = i;
    for (int i = 0; i < n; i++)
      if (p[i] == -1) nr[ucnt++] = i;
    if (!ncnt) break;
    int s = rand() % ucnt;
    int i = nr[s];
    for (int j = 0; j < ncnt; j++) {
      int ed = nu[j] ^ a[i];
      if (pl[ed] == -1) {
        p[i] = nu[j], q[i] = ed;
        break;
      }
    }
    if (p[i] == -1) {
      int nid = rand() % ncnt;
      p[i] = nu[nid];
      int ed = nu[nid] ^ a[i];
      p[pl[ed]] = q[pl[ed]] = -1;
      q[i] = ed;
    }
  }
  cout << "Shi" << endl;
  for (int i = 0; i < n; i++) cout << p[i] << " ";
  cout << endl;
  for (int i = 0; i < n; i++) cout << q[i] << " ";
  cout << endl;
  return 0;
}