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>
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  int k;
  std::cin >> k;
  std::vector<int> target(1 << k);
  int xorall = 0;
  for (int& ai : target) {
    std::cin >> ai;
    xorall ^= ai;
  }
  if (xorall != 0) {
    std::cout << "Fou\n";
    return 0;
  }
  std::vector<int> p(1 << k), q(1 << k), a(1 << k), ppos(1 << k);
  for (int i = 1 << k; i--;) p[i] = q[i] = ppos[i] = i;
  auto swapp = [&](int i, int j) {
    std::swap(p[i], p[j]);
    ppos[p[i]] = i;
    ppos[p[j]] = j;
  };
  for (int i = 1; i < (1 << k); ++i) {
    int x = a[i - 1] xor target[i - 1];
    if (x) {
      a[i - 1] ^= x;
      a[i] ^= x;
      int m = i, n = i - 1;
      for (;;) {
        if ((p[m] xor q[m]) == a[m]) break;
        if ((p[n] xor q[m]) == a[m]) {
          swapp(m, n);
          assert((p[m] xor q[m]) == a[m]);
          assert((p[n] xor q[n]) == a[n]);
          break;
        }
        int t = ppos[a[m] xor q[m]];
        assert(t != m and t != n);
        swapp(m, t);
        assert((p[m] xor q[m]) == a[m]);
        m = t;
        std::swap(q[m], q[n]);
      }
    }
    assert(std::equal(begin(a), begin(a) + i, begin(target)));
  }
  assert(a == target);
  std::cout << "Shi\n";
  for (int x : p) std::cout << x << ' ';
  std::cout << '\n';
  for (int x : q) std::cout << x << ' ';
  std::cout << '\n';
}