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 re_ad() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return x * f;
}
int n, k;
int a[5010], sum;
int p[5010], q[5010];
bool fla[5010];
int pla[5010];
int main() {
  register int i, x, y, j, z, ii;
  bool fl = false;
  srand(time(0));
  k = re_ad();
  n = 1 << k;
  for (i = 0; i < n; ++i) {
    a[i] = re_ad();
    sum ^= a[i];
  }
  if (sum) {
    puts("Fou");
    return 0;
  }
  memset(p, -1, sizeof(p));
  memset(q, -1, sizeof(q));
  memset(pla, -1, sizeof(p));
  i = 0;
  while (i < n) {
    x = rand() & (n - 1);
    while (p[x] >= 0) x = rand() & (n - 1);
    fl = false;
    for (j = 0; j < n; ++j) {
      if (!fla[j] && pla[j ^ a[x]] == -1) {
        p[x] = j;
        q[x] = j ^ a[x];
        fl = true;
        fla[j] = true;
        pla[j ^ a[x]] = x;
        break;
      }
    }
    if (fl) {
      ++i;
      continue;
    }
    y = rand() & (n - 1);
    while (fla[y]) y = rand() & (n - 1);
    z = pla[y ^ a[x]];
    p[x] = y;
    q[x] = y ^ a[x];
    fla[y] = true;
    pla[y ^ a[x]] = x;
    fla[p[z]] = false;
    q[z] = p[z] = -1;
  }
  puts("Shi");
  for (i = 0; i < n; ++i) printf("%d ", p[i]);
  puts("");
  for (i = 0; i < n; ++i) printf("%d ", q[i]);
  return 0;
}