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>
#pragma warning(disable : 4996)
int __builtin_popcount(int n) {
  int x = 0;
  while (n) n -= n & (-n), x++;
  return x;
}
using namespace std;
const int mod = 1000000009;
const int inf = 2012345678;
const long long infl = 9012345678901234567;
const double pi = 3.1415926535897932384626433832795;
const int N = (1 << 12) + 10;
int a[N], s[N], t[N], rt[N], b[N], c[N], chk[N], se[N], m;
int main() {
  freopen("input.txt", "r", stdin);
  int k;
  scanf("%d", &(k));
  for (int i = 0; i < (1 << k); i++) scanf("%d", &(a[i]));
  int X = 0;
  for (int i = 0; i < (1 << k); i++) X ^= a[i];
  if (X) return puts("Fou"), 0;
  puts("Shi");
  for (int i = 0; i < (1 << k); i++) s[i] = t[i] = i;
  for (int i = 0; i < (1 << k); i++) {
    if (a[i] == b[i]) continue;
    for (int j = 0; j < (1 << k); j++) rt[t[j]] = j, c[j] = b[j], chk[j] = 0;
    m = 2;
    se[0] = i;
    se[1] = i + 1;
    chk[i] = chk[i + 1] = 1;
    int X = a[i] ^ b[i];
    c[i] ^= X;
    c[i + 1] ^= X;
    int p = 0;
    while (1) {
      int j = rt[c[se[m - 1]] ^ s[se[m - 2]]];
      se[m] = j;
      if (chk[j]) {
        if (j == se[1]) p = 1;
        break;
      }
      chk[j] = 1;
      m++;
    }
    int ss = s[se[m - 1]];
    for (int i = m; --i;) s[se[i]] = s[se[i - 1]];
    s[se[0]] = ss;
    ss = t[se[p ^ 1]];
    int tt = t[se[p]];
    for (int j = 1; j < m; j++) t[se[j]] = t[se[j + 1]];
    if (p == 1) t[se[m - 1]] = tt;
    t[se[0]] = ss;
    b[i] = c[i];
    b[i + 1] = c[i + 1];
  }
  for (int i = 0; i < (1 << k); i++) printf("%d ", t[i]);
  puts("");
  for (int i = 0; i < (1 << k); i++) printf("%d ", s[i]);
  puts("");
  return 0;
}