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 = 4105;
int n, a[maxn], b[maxn], c[maxn], d[maxn], plc[maxn], plb[maxn];
void fix(int x, int y) {
  if ((b[x] ^ c[x]) == a[x] && (b[y] ^ c[y]) == a[y]) return;
  if ((b[y] ^ c[y]) == a[x] && (b[x] ^ c[x]) == a[y]) {
    swap(plb[b[x]], plb[b[y]]);
    swap(plc[c[x]], plc[c[y]]);
    swap(b[x], b[y]);
    swap(c[x], c[y]);
    return;
  }
  int p = plc[a[x] ^ b[x]];
  swap(plc[a[x] ^ b[x]], plc[c[x]]);
  swap(c[x], c[p]);
  swap(plb[b[y]], plb[a[x] ^ b[x]]);
  swap(b[y], b[p]);
  fix(p, y);
}
int main() {
  scanf("%d", &n);
  n = (1 << n);
  int x = 0;
  for (int i = 1; i <= n; i++)
    scanf("%d", &d[i]), x ^= d[i], c[i] = b[i] = i - 1, plc[i - 1] = i,
                                   plb[i - 1] = i;
  if (x != 0) {
    puts("Fou");
    return 0;
  }
  puts("Shi");
  for (int i = 1; i < n; i++) {
    int p = d[i] ^ a[i];
    a[i] ^= p, a[i + 1] ^= p;
    fix(i, i + 1);
  }
  for (int i = 1; i <= n; i++) printf("%d ", b[i]);
  puts("");
  for (int i = 1; i <= n; i++) printf("%d ", c[i]);
  puts("");
}