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) + 10;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int a[N], b[N], p[N], q[N], loc[N];
void upd(int i, int j, int x);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int k;
  cin >> k;
  int n = (1 << k);
  int test = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    p[i] = q[i] = i;
    test ^= a[i];
  }
  if (test) {
    cout << "Fou\n";
    return 0;
  }
  shuffle(p, p + n, rng);
  shuffle(q, q + n, rng);
  for (int i = 0; i < n; ++i) {
    b[i] = p[i] ^ q[i];
    loc[p[i]] = i;
  }
  cout << "Shi\n";
  for (int i = 0; i < n - 1; ++i) {
    upd(i, i + 1, b[i] ^ a[i]);
  }
  for (int i = 0; i < n; ++i) {
    cout << p[i] << ' ';
  }
  cout << '\n';
  for (int i = 0; i < n; ++i) {
    cout << q[i] << ' ';
  }
  cout << '\n';
}
void upd(int i, int j, int x) {
  int s = i;
  b[i] ^= x;
  b[j] ^= x;
  while ((p[i] ^ q[i]) != b[i]) {
    int t = loc[b[i] ^ q[i]];
    swap(loc[p[i]], loc[p[t]]);
    swap(p[i], p[t]);
    swap(q[t], q[j]);
    i = t;
    if (i > s) {
      break;
    }
  }
}