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;
template <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
template <class T>
T gcd(T a, T b) {
  return !b ? a : gcd(b, a % b);
}
template <class T>
inline void umin(T &x, T y) {
  x = x < y ? x : y;
}
template <class T>
inline void umax(T &x, T y) {
  x = x > y ? x : y;
}
inline unsigned int R() {
  static unsigned int seed = 416;
  return seed ^= seed >> 5, seed ^= seed << 17, seed ^= seed >> 13;
}
const int N = 1 << 12 | 3;
int k, n, a[N], p[N], q[N];
int main() {
  read(k);
  n = 1 << k;
  int s = 0;
  for (register int i = (0); i <= (n - 1); i++)
    read(a[i]), s ^= a[i], p[i] = q[i] = -1;
  if (s) {
    puts("Fou");
    return 0;
  }
  while (1) {
    static int pval[N], qpos[N];
    for (register int i = (0); i <= (n - 1); i++) pval[i] = 0, qpos[i] = -1;
    for (register int i = (0); i <= (n - 1); i++)
      if (p[i] != -1) pval[p[i]] = 1;
    for (register int i = (0); i <= (n - 1); i++)
      if (q[i] != -1) qpos[q[i]] = i;
    static int sp[N], val[N];
    int sz = 0;
    for (register int i = (0); i <= (n - 1); i++)
      if (p[i] == -1) sp[sz++] = i;
    sz = 0;
    for (register int i = (0); i <= (n - 1); i++)
      if (!pval[i]) val[sz++] = i;
    if (!sz) break;
    int x = sp[R() % sz];
    for (register int i = (0); i <= (sz - 1); i++)
      if (qpos[a[x] ^ val[i]] == -1) {
        p[x] = val[i];
        q[x] = a[x] ^ val[i];
        break;
      }
    if (p[x] == -1) {
      p[x] = val[R() % sz];
      q[x] = a[x] ^ p[x];
      int y = qpos[q[x]];
      assert(q[y] == q[x]);
      q[y] = p[y] = -1;
    }
  }
  puts("Shi");
  for (register int i = (0); i <= (n - 1); i++) printf("%d ", p[i]);
  puts("");
  for (register int i = (0); i <= (n - 1); i++) printf("%d ", q[i]);
  puts("");
  return 0;
}