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 x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
void wr(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) wr(x / 10);
  putchar('0' + x % 10);
}
mt19937 gen(20040502);
const int MAXN = 4100;
int n, a[MAXN], A[MAXN], B[MAXN], pa[MAXN], pb[MAXN], tmp[MAXN], cnt,
    tmp2[MAXN], tot;
int main() {
  n = 1 << read();
  for (register int(i) = (1); (i) <= (n); ++(i)) a[i] = read();
  memset(A, -1, sizeof A);
  memset(B, -1, sizeof B);
  while (clock() / CLOCKS_PER_SEC < 0.95) {
    memset((pa), 0, sizeof(pa));
    memset((pb), 0, sizeof(pb));
    cnt = tot = 0;
    for (register int(i) = (1); (i) <= (n); ++(i))
      if (A[i] != -1)
        pa[A[i]] = i, pb[B[i]] = i;
      else
        tmp[++cnt] = i;
    if (!cnt) {
      puts("Shi");
      for (register int(i) = (1); (i) <= (n); ++(i)) wr(A[i]), putchar(' ');
      puts("");
      for (register int(i) = (1); (i) <= (n); ++(i)) wr(B[i]), putchar(' ');
      puts("");
      return 0;
    }
    for (register int(i) = (0); (i) <= (n - 1); ++(i))
      if (!pa[i]) tmp2[++tot] = i;
    int p = tmp[gen() % cnt + 1], v, t;
    for (register int(i) = (1); (i) <= (tot); ++(i)) {
      int t = a[p] ^ tmp2[i];
      if (!pb[t]) {
        A[p] = tmp2[i];
        B[p] = t;
        goto ss;
      }
    }
    v = tmp2[gen() % tot + 1];
    t = a[p] ^ v;
    A[pb[t]] = B[pb[t]] = -1;
    B[p] = t;
    A[p] = v;
  ss:;
  }
  puts("Fou");
  return 0;
}