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>
const int N = 1 << 12;
int k, n, a[N], p[N], q[N], r[N];
void do_xor(int x, int y, int z) {
  int t = r[p[x] ^ z];
  if (t == x) return;
  if (t == y) {
    std::swap(p[x], p[y]);
    r[p[x]] = x, r[p[y]] = y;
    return;
  }
  std::swap(p[x], p[t]);
  r[p[x]] = x, r[p[t]] = t;
  std::swap(q[y], q[t]);
  do_xor(t, y, z ^ q[y] ^ q[t]);
}
int main() {
  assert(scanf("%d", &k) == 1);
  n = 1 << k;
  for (int i = 0; i < n; ++i) assert(scanf("%d", &a[i]) == 1);
  for (int i = 0; i < n; ++i) p[i] = q[i] = r[i] = i;
  for (int i = n; i--;) {
    do_xor(i - 1, i, a[i]);
    a[i - 1] ^= a[i];
  }
  if (a[0] == 0) {
    puts("Shi");
    for (int i = 0; i < n; ++i) printf("%d%c", p[i], i + 1 == n ? '\n' : ' ');
    for (int i = 0; i < n; ++i) printf("%d%c", q[i], i + 1 == n ? '\n' : ' ');
  } else {
    puts("Fou");
  }
  return 0;
}