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;
const int N = 5000 + 5;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, a[N];
bool vis1[N], vis2[N];
int lef[N], rig[N], re[N], ans[N];
inline int random(int l, int r) {
  return (long long)rand() * rand() % (r - l + 1) + l;
}
int main() {
  srand(time(0));
  n = read();
  int val = 0;
  for (int i = 0; i < (1 << n); i++) a[i] = read(), val ^= a[i];
  if (val) {
    printf("Fou\n");
    return 0;
  }
  printf("Shi\n");
  int cnt = 0;
  memset(lef, -1, sizeof(lef));
  memset(rig, -1, sizeof(rig));
  while (cnt < (1 << n)) {
    int len = 0;
    for (int i = 0; i < (1 << n); i++)
      if (!vis1[i]) re[++len] = i;
    int tmp = random(1, len);
    tmp = re[tmp];
    vis1[tmp] = true;
    bool flag = false;
    for (int i = 0; i < (1 << n); i++) {
      int val = (a[i] ^ tmp);
      if (!vis2[i] && lef[val] == -1) {
        lef[val] = tmp;
        rig[val] = i;
        cnt++;
        vis2[i] = true;
        ans[i] = tmp;
        flag = true;
        break;
      }
    }
    if (flag) continue;
    len = 0;
    for (int i = 0; i < (1 << n); i++)
      if (!vis2[i]) re[++len] = i;
    int p = random(1, len);
    p = re[p];
    vis2[p] = true;
    int val = (a[p] ^ tmp);
    vis1[lef[val]] = false;
    vis2[rig[val]] = false;
    lef[val] = tmp;
    rig[val] = p;
    ans[p] = tmp;
  }
  for (int i = 0; i < (1 << n); i++) printf("%d ", ans[i]);
  printf("\n");
  for (int i = 0; i < (1 << n); i++) printf("%d ", a[i] ^ ans[i]);
  printf("\n");
  return 0;
}