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 = 1e6;
namespace xor_permutation {
const int N = 1 << 20;
int a[N], b[N], f[N];
int ind[N];
void fix(int x, int y) {
  if (f[x] == (a[x] ^ b[x])) {
    return;
  }
  if (f[x] == (a[x] ^ b[y])) {
    swap(b[y], b[x]);
    swap(ind[b[x]], ind[b[y]]);
    return;
  }
  if (f[x] == (a[y] ^ b[y])) {
    swap(a[x], a[y]);
    swap(b[x], b[y]);
    swap(ind[b[x]], ind[b[y]]);
    return;
  }
  if (f[x] == (a[y] ^ b[x])) {
    swap(a[x], a[y]);
    return;
  }
  int z = ind[f[x] ^ a[x]];
  swap(b[z], b[x]);
  swap(ind[b[z]], ind[b[x]]);
  swap(a[y], a[z]);
  fix(z, y);
}
bool work(int* c, int k) {
  int all = 0;
  for (int i = 0; i < (1 << k); i++) {
    a[i] = b[i] = ind[i] = i;
    f[i] = 0;
    all ^= c[i];
  }
  if (all) return 0;
  for (int i = 0; i < (1 << k) - 1; i++) {
    f[i + 1] ^= (f[i] ^ c[i]);
    f[i] = c[i];
    fix(i, i + 1);
  }
  return 1;
}
}  // namespace xor_permutation
int a[1000000];
int main() {
  int k;
  scanf("%d", &k);
  for (int i = 0; i < (1 << k); i++) {
    scanf("%d", &a[i]);
  }
  bool isans = xor_permutation::work(a, k);
  if (isans) {
    printf("Shi\n");
    for (int i = 0; i < (1 << k); i++) printf("%d ", xor_permutation::a[i]);
    printf("\n");
    for (int i = 0; i < (1 << k); i++) printf("%d ", xor_permutation::b[i]);
    printf("\n");
  } else
    printf("Fou");
  return 0;
}