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 n, A[4205];
int Lx[4205], Ly[4205], V[4205];
int Ux[4205], Uy[4205];
int stk[4205], top;
void Read(int &p) {
  p = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) p = p * 10 + c - '0';
}
int main() {
  Read(n), n = 1 << n;
  int tim = clock(), s = 0;
  for (int i = 0; i < n; i++) {
    Read(A[i]), s ^= A[i];
    Ux[i] = Uy[i] = -1;
  }
  if (s) {
    puts("Fou");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (!V[i ^ A[i]]) {
      V[i ^ A[i]] = 1;
      Lx[i ^ A[i]] = Ly[i ^ A[i]] = i;
      Ux[i] = Uy[i] = i;
    }
  }
  const int lim = CLOCKS_PER_SEC * 0.95;
  while (clock() - tim < lim) {
    top = 0;
    for (int i = 0; i < n; i++)
      if (Ux[i] == -1) stk[++top] = i;
    if (!top) break;
    int x = stk[rand() % top + 1];
    top = 0;
    for (int i = 0; i < n; i++)
      if (Uy[i] == -1 && !V[x ^ A[i]]) stk[++top] = i;
    if (top) {
      int y = stk[rand() % top + 1];
      V[x ^ A[y]] = 1;
      Lx[x ^ A[y]] = x, Ly[x ^ A[y]] = y;
      Ux[x] = y, Uy[y] = x;
    } else {
      top = 0;
      for (int i = 0; i < n; i++)
        if (Uy[i] == -1) stk[++top] = i;
      int y = stk[rand() % top + 1];
      Ux[Lx[x ^ A[y]]] = Uy[Ly[x ^ A[y]]] = -1;
      Lx[x ^ A[y]] = x, Ly[x ^ A[y]] = y;
      Ux[x] = y, Uy[y] = x;
    }
  }
  puts("Shi");
  for (int i = 0; i < n; i++) printf("%d%c", Uy[i], i == n - 1 ? 10 : 32);
  for (int i = 0; i < n; i++)
    printf("%d%c", A[i] ^ Uy[i], i == n - 1 ? 10 : 32);
}