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;
int read() {
  char ch = getchar();
  int x = 0, fl = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fl = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch - '0');
  return x * fl;
}
const int N = 4205;
int k, n, xor_all;
int a[N], p[N], q[N], rev[N], pos[N], num[N];
bool visp[N], visq[N];
int main() {
  srand(19950910);
  k = read();
  n = (1 << k);
  for (int i = 1; i <= n; i++) a[i] = read(), xor_all ^= a[i];
  if (xor_all) {
    puts("Fou");
    return 0;
  }
  memset(p, -1, sizeof(p));
  memset(q, -1, sizeof(q));
  while (1) {
    int topp = 0, topn = 0;
    for (int i = 1; i <= n; i++)
      if (p[i] == -1) pos[++topp] = i;
    for (int i = 1; i <= n; i++)
      if (~q[i]) rev[q[i]] = i;
    memset(visp, 0, sizeof(visp));
    memset(visq, 0, sizeof(visq));
    for (int i = 1; i <= n; i++) visp[p[i]] = visq[q[i]] = true;
    if (!topp) break;
    for (int i = 0; i < n; i++)
      if (!visp[i]) num[++topn] = i;
    int pv = pos[rand() % topp + 1];
    bool ff = 0;
    for (int i = 1; i <= topn; i++) {
      int tmp = num[i] ^ a[pv];
      if (!visq[tmp]) {
        p[pv] = num[i], q[pv] = tmp;
        ff = 1;
        break;
      }
    }
    if (ff) continue;
    int nv = num[rand() % topn + 1], tmp = nv ^ a[pv];
    p[pv] = nv;
    q[pv] = tmp;
    p[rev[tmp]] = q[rev[tmp]] = -1;
  }
  puts("Shi");
  for (int i = 1; i <= n; i++) printf("%d ", p[i]);
  puts("");
  for (int i = 1; i <= n; i++) printf("%d ", q[i]);
  puts("");
  return 0;
}