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;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1e5 + 3;
const int infint = (int)1e9 + 3;
const long long inf = (long long)2e18;
int n, a[MAXN], p[MAXN], q[MAXN], b[MAXN], pos[MAXN];
void print() {
  for (int i = 0; i < n; i++) cout << p[i] << " ";
  cout << "\n";
  for (int i = 0; i < n; i++) cout << q[i] << " ";
  cout << "\n";
  for (int i = 0; i < n; i++) cout << pos[i] << " ";
  cout << "\n";
  cout << "\n";
}
inline void make(int i, int j) {
  int t = pos[q[i] ^ b[i]];
  if (t == i) return;
  if (t == j) {
    swap(p[i], p[j]);
    pos[p[i]] = i, pos[p[j]] = j;
    return;
  }
  int pi = p[i], qi = q[i], pt = p[t], qt = q[t], pj = p[j], qj = q[j];
  p[t] = pi, q[t] = qj;
  p[i] = pt;
  q[i] = qi;
  p[j] = pj;
  q[j] = qt;
  pos[p[t]] = t, pos[p[i]] = i, pos[p[j]] = j;
  make(t, j);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  n = (1 << n);
  int mika = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mika ^= a[i];
  }
  if (mika) return cout << "Fou", 0;
  cout << "Shi"
       << "\n";
  for (int i = 0; i < n; i++) p[i] = i, q[i] = i;
  for (int i = 0; i < n; i++) pos[i] = i;
  for (int i = 0; i < n - 1; i++)
    b[i + 1] ^= a[i] ^ b[i], b[i] = a[i], make(i, i + 1);
  for (int i = 0; i < n; i++) cout << p[i] << " ";
  cout << "\n";
  for (int i = 0; i < n; i++) cout << q[i] << " ";
}