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;
long long int mod = 998244353;
const long double error = 1e-7;
const double PI = acos(-1);
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
inline long long int MOD(long long int x, long long int m = mod) {
  long long int y = x % m;
  return (y >= 0) ? y : y + m;
}
const int inf = 1e9;
const long long int infl = 1e18;
const int nmax = (1LL << (12)) + 10;
int A[nmax], B[nmax];
int invA[nmax];
int P[nmax];
int input[nmax];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  n = (1LL << (n));
  int xorval = 0;
  for (int i = 0; i < n; i++) {
    cin >> input[i];
    xorval ^= input[i];
    A[i] = i, invA[i] = i;
    B[i] = i;
    P[i] = 0;
  }
  if (xorval != 0) {
    cout << "Fou\n";
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    if (P[i] == input[i]) continue;
    int x = input[i] ^ P[i];
    P[i] ^= x;
    P[n - 1] ^= x;
    int j = i;
    while (1) {
      int a = P[j] ^ B[j];
      int k = invA[a];
      swap(A[j], A[k]);
      invA[A[j]] = j;
      invA[A[k]] = k;
      if ((A[k] ^ B[k]) == P[k]) break;
      swap(B[k], B[n - 1]);
      j = k;
    }
  }
  cout << "Shi\n";
  for (int i = 0; i < n; i++) cout << A[i] << " ";
  cout << "\n";
  for (int i = 0; i < n; i++) cout << B[i] << " ";
  cout << "\n";
  return 0;
}