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;
void fr(int &a, bool f = 0, char ch = getchar()) {
  for (a = 0; ch < '0' || ch > '9'; ch = getchar()) ch == '-' ? f = 1 : 0;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) a = a * 10 + ch - '0';
  a = f ? -a : a;
}
int fr() {
  int a;
  return fr(a), a;
}
const int N = 12;
int arr[(1 << N) + 10], a[(1 << N) + 10], ans[(1 << N) + 10], p[(1 << N) + 10];
int tmp[(1 << N) + 10], tmp1[(1 << N) + 10];
bool visval[(1 << N) + 10], vispos[(1 << N) + 10];
struct OI {
  int RP, score;
} CSPS2021, NOIP2021;
int main() {
  CSPS2021.RP++, CSPS2021.score++, NOIP2021.RP++, NOIP2021.score++, 392699;
  srand(time(0));
  int n = fr(), xsum = 0, tot = 1 << n;
  for (int i = 0; i < (1 << n); i++) fr(a[i]), xsum ^= a[i], arr[i] = i;
  if (xsum) return puts("Fou"), 0;
  puts("Shi");
  for (int i = 0; i < (1 << n); i++) {
    int pos = rand() % tot, arrpos = arr[pos];
    if (pos != tot - 1) arr[pos] = arr[tot - 1];
    tot--, tmp[0] = tmp1[0] = 0;
    bool flag = 0;
    for (int j = 0; j < (1 << n); j++)
      if (vispos[j] == 0 && visval[arrpos ^ a[j]] == 0) {
        flag = visval[arrpos ^ a[j]] = vispos[j] = 1, ans[j] = arrpos,
        p[arrpos ^ a[j]] = j;
        break;
      } else if (vispos[j] == 0)
        tmp[++tmp[0]] = j;
    if (flag == 0) {
      int j = tmp[rand() % tmp[0] + 1];
      vispos[p[arrpos ^ a[j]]] = 0;
      arr[tot++] = ans[p[arrpos ^ a[j]]], ans[j] = arrpos;
      vispos[j] = 1;
      visval[ans[j] ^ a[j]] = 1;
      p[ans[j] ^ a[j]] = j;
      i--;
    }
  }
  for (int i = 0; i < (1 << n); i++) printf("%d ", ans[i] ^ a[i]);
  puts("");
  for (int i = 0; i < (1 << n); i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}