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 MAXN = 4096;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
int k, n, p[MAXN], q[MAXN], t[MAXN], a[MAXN], home[MAXN];
void fix(int i, int j) {
  int k = home[t[i] ^ q[i]];
  if (k == i) return;
  if (k == j) {
    swap(home[p[i]], home[p[j]]);
    swap(p[i], p[j]);
    return;
  }
  swap(home[p[i]], home[p[k]]);
  swap(p[i], p[k]), swap(q[j], q[k]);
  fix(k, j);
}
int main() {
  read(k), n = 1 << k;
  int sum = 0;
  for (int i = 0; i < n; i++) {
    read(a[i]);
    sum ^= a[i];
    p[i] = q[i] = home[i] = i;
  }
  if (sum != 0) {
    puts("Fou");
    return 0;
  }
  for (int i = 0; i < n - 1; i++)
    if (a[i] != t[i]) {
      int delta = t[i] ^ a[i];
      t[i] ^= delta, t[i + 1] ^= delta;
      fix(i, i + 1);
    }
  puts("Shi");
  for (int i = 0; i < n; i++) printf("%d ", p[i]);
  printf("\n");
  for (int i = 0; i < n; i++) printf("%d ", q[i]);
  printf("\n");
  return 0;
}