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 N = (1 << 12);
int k, n, s, p[N], q[N], a[N], b[N], iq[N];
void add(int i, int x) {
  if (x == 0) return;
  int p1 = p[i], q1 = q[i], q2 = q[i + 1];
  a[i] ^= x;
  a[i + 1] ^= x;
  int j = i + 1;
  while ((p1 ^ a[j]) != q1 && (p1 ^ a[j]) != q2) {
    int k = iq[p1 ^ a[j]];
    swap(p[j], p1);
    q[j] = p[j] ^ a[j];
    iq[q[j]] = j;
    j = k;
  }
  swap(p1, p[j]);
  swap(p1, p[i]);
  if ((p[j] ^ a[j]) == q1)
    q[j] = q1, iq[q1] = j, q[i] = q2, iq[q2] = i;
  else
    q[j] = q2, iq[q2] = j, q[i] = q1, iq[q1] = i;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> k;
  n = 1 << k;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    s ^= b[i];
  }
  if (s != 0) {
    cout << "Fou\n";
    return 0;
  }
  for (int i = 0; i < n; i++) p[i] = q[i] = iq[i] = i;
  for (int i = 0; i < n - 1; i++) add(i, a[i] ^ b[i]);
  cout << "Shi\n";
  for (int i = 0; i < n; i++) cout << p[i] << (i == n - 1 ? "\n" : " ");
  for (int i = 0; i < n; i++) cout << q[i] << (i == n - 1 ? "\n" : " ");
  return 0;
}