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;
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - 48;
    s = getchar();
  }
  return f * x;
}
const int N = 5005;
int id1[N], id2[N], p[N], q[N], a[N];
int tmp1[N], tmp2[N];
int main() {
  int k = read(), n = 1 << k;
  for (int i = 1; i <= n; i++) a[i] = read();
  memset(p, -1, sizeof(p)), memset(q, -1, sizeof(q));
  while (clock() / CLOCKS_PER_SEC < 0.8) {
    int tot = 0, cnt = 0;
    memset(id1, 0, sizeof(id1)), memset(id2, 0, sizeof(id2));
    for (int i = 1; i <= n; i++)
      if (p[i] != -1)
        id1[p[i]] = i, id2[q[i]] = i;
      else
        tmp1[++cnt] = i;
    if (!cnt) {
      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;
    }
    for (int i = 0; i < n; i++)
      if (!id1[i]) tmp2[++tot] = i;
    int res = tmp1[rand() % cnt + 1];
    int flag = 1;
    for (int i = 1; i <= tot; i++) {
      int t = a[res] ^ tmp2[i];
      if (!id2[t]) {
        p[res] = tmp2[i];
        q[res] = t;
        flag = 0;
        break;
      }
    }
    if (flag) {
      int v = tmp2[rand() % tot + 1];
      int t = a[res] ^ v;
      p[id2[t]] = q[id2[t]] = -1;
      p[res] = v, q[res] = t;
    }
  }
  puts("Fou");
  return 0;
}