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;
template <typename T>
vector<T>& operator--(vector<T>& v) {
  for (auto& i : v) --i;
  return v;
}
template <typename T>
vector<T>& operator++(vector<T>& v) {
  for (auto& i : v) ++i;
  return v;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  for (auto& i : v) os << i << ' ';
  return os;
}
template <typename T, typename U>
istream& operator>>(istream& is, pair<T, U>& p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& p) {
  os << p.first << ' ' << p.second;
  return os;
}
template <typename T, typename U>
pair<T, U> operator-(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <typename T, typename U>
pair<T, U> operator+(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
template <typename T, typename U>
void umin(T& a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
void umax(T& a, U b) {
  if (a < b) a = b;
}
int main() {
  int k;
  cin >> k;
  int n = 1 << k;
  vector<int> a(n);
  cin >> a;
  vector<int> p, q;
  vector<int> ind;
  int u = 0;
  for (int i = 0; i < n; ++i) {
    u ^= a[i];
    p.push_back(i);
    q.push_back(i);
    ind.push_back(i);
  }
  if (u != 0) {
    cout << "Fou\n";
    return 0;
  }
  cout << "Shi\n";
  vector<int> b(n, 0);
  function<void(int, int)> fix = [&](int i, int j) {
    42;
    ;
    if (b[i] == (p[i] ^ q[i])) {
      return;
    }
    if (b[i] == (p[i] ^ q[j])) {
      swap(q[i], q[j]);
      return;
    }
    if (b[i] == (p[j] ^ q[i])) {
      swap(ind[p[i]], ind[p[j]]);
      swap(p[i], p[j]);
      return;
    }
    if (b[i] == (p[j] ^ q[j])) {
      swap(q[i], q[j]);
      swap(p[i], p[j]);
      swap(ind[p[i]], ind[p[j]]);
      return;
    }
    int t = ind[b[i] ^ q[i]];
    swap(ind[p[i]], ind[p[t]]);
    swap(p[i], p[t]);
    swap(q[t], q[j]);
    fix(t, j);
  };
  for (int i = 0; i < n - 1; ++i) {
    42;
    ;
    int d = (a[i] ^ b[i]);
    b[i] ^= d;
    b[i + 1] ^= d;
    fix(i, i + 1);
  }
  cout << p << '\n';
  cout << q << '\n';
  return 0;
}