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 qr() {
  int x = 0, f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int N = 13;
int n;
int sum, x[1 << N], match[1 << N], st[1 << N], top, sp[1 << N], tp;
int o[1 << N], p[1 << N];
int main() {
  srand(time(0));
  n = qr();
  for (int i = 0; i < 1 << n; i++) {
    x[i] = qr(), sum ^= x[i];
  }
  if (sum) {
    puts("Fou");
    return 0;
  } else {
    puts("Shi");
    int len = 1 << n;
    memset(match, -1, sizeof match);
    while (1) {
      memset(o, 0, sizeof o);
      memset(p, -1, sizeof p);
      for (int i = 0; i < len; i++) {
        if (match[i] != -1) o[match[i]] = 1, p[match[i] ^ x[i]] = i;
      }
      top = 0, tp = 0;
      for (int i = 0; i < len; i++) {
        if (!o[i]) sp[tp++] = i;
      }
      for (int i = 0; i < len; i++) {
        if (match[i] == -1) st[top++] = i;
      }
      if (!tp) break;
      int v = rand() % top;
      int i = st[v];
      for (int j = 0; j < tp; j++) {
        int z = sp[j] ^ x[i];
        if (p[z] == -1) {
          match[i] = sp[j];
          break;
        }
      }
      if (match[i] == -1) {
        int u = rand() % tp;
        match[i] = sp[u];
        int z = x[i] ^ sp[u];
        match[p[z]] = -1;
        p[z] = i;
      }
    }
    for (int i = 0; i < len; i++) {
      printf("%d ", match[i]);
    }
    printf("\n");
    for (int i = 0; i < len; i++) {
      printf("%d ", match[i] ^ x[i]);
    }
  }
  return 0;
}