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;
inline int read() {
  int f = 1, x = 0;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f * x;
}
int k;
int n;
int a[5010];
int val[5010];
int id[5010];
int p[5010], q[5010];
long long sum;
inline void solve(int l, int r) {
  int k = id[val[l] ^ q[l]];
  if (k == l) return;
  if (k == r) {
    swap(id[p[l]], id[p[k]]);
    swap(p[l], p[k]);
    return;
  } else {
    swap(id[p[l]], id[p[k]]);
    swap(p[l], p[k]);
    swap(q[r], q[k]);
    solve(k, r);
  }
}
int main() {
  k = read();
  n = (1 << k);
  for (int i = 0; i < n; i++)
    a[i] = read(), sum ^= a[i], id[i] = p[i] = q[i] = i, val[i] = 0;
  if (sum) {
    cout << "Fou" << endl;
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    if (a[i] ^ val[i]) {
      val[i + 1] = val[i + 1] ^ (val[i] ^ a[i]);
      val[i] = a[i];
      solve(i, i + 1);
    }
  }
  cout << "Shi" << endl;
  for (int i = 0; i < n; i++) cout << p[i] << " ";
  cout << endl;
  for (int i = 0; i < n; i++) cout << q[i] << " ";
  return 0;
}