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 rlen = 1 << 18 | 1;
inline char gc() {
  static char buf[rlen], *ib, *ob;
  (ib == ob) && (ob = (ib = buf) + fread(buf, 1, rlen, stdin));
  return ib == ob ? -1 : *ib++;
}
inline int read() {
  int ans = 0;
  char ch = gc();
  while (!isdigit(ch)) ch = gc();
  while (isdigit(ch)) ans = ((ans << 2) + ans << 1) + (ch ^ 48), ch = gc();
  return ans;
}
const int N = 5005;
int k, a[N], n, p[N], q[N], idx[N], vl[N];
inline void fix(int x, int y) {
  int k = idx[vl[x] ^ q[x]];
  if (k == x) return;
  if (k == y) {
    swap(idx[p[x]], idx[p[y]]);
    swap(p[x], p[y]);
    return;
  }
  swap(idx[p[x]], idx[p[k]]);
  swap(p[x], p[k]), swap(q[y], q[k]);
  fix(k, y);
}
int main() {
  n = 1 << (k = read());
  int ss = 0;
  for (register int i = 0; i < n; ++i)
    ss ^= (a[i] = read()), p[i] = q[i] = idx[i] = i;
  if (ss) return puts("Fou"), 0;
  puts("Shi");
  for (register int i = 0, dt; i < n - 1; ++i) {
    if (a[i] ^ vl[i]) {
      vl[i + 1] ^= vl[i] ^ a[i];
      vl[i] = a[i];
      fix(i, i + 1);
    }
  }
  for (register int i = 0; i < n; ++i) cout << p[i] << ' ';
  puts("");
  for (register int i = 0; i < n; ++i) cout << q[i] << ' ';
  return 0;
}