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 readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, ncnt, cnt;
int a[5005], p[5005], q[5005], dp[5005], dq[5005], tmp[5005], emp[5005];
int main() {
  n = 1 << readint();
  int sum = 0;
  for (int i = 1; i <= n; i++) a[i] = readint(), sum ^= a[i];
  if (sum) return printf("Fou\n"), 0;
  memset(p, -1, sizeof(p));
  memset(q, -1, sizeof(q));
  while (1) {
    cnt = 0;
    for (int i = 1; i <= n; i++)
      if (p[i] < 0) emp[++cnt] = i;
    if (!cnt) break;
    int now = emp[rand() % cnt + 1];
    for (int i = 0; i < n; i++) {
      if (!dp[i] && !dq[a[now] ^ i]) {
        p[now] = i, q[now] = a[now] ^ i, dp[i] = now, dq[a[now] ^ i] = now;
        break;
      }
    }
    if (p[now] < 0) {
      ncnt = 0;
      for (int i = 0; i < n; i++)
        if (!dp[i]) tmp[++ncnt] = i;
      int th = rand() % ncnt + 1, g = a[now] ^ tmp[th];
      p[now] = tmp[th], dp[tmp[th]] = now, q[now] = g, dp[p[dq[g]]] = 0,
      p[dq[g]] = -1, q[dq[g]] = -1, dq[g] = now;
    }
  }
  printf("Shi\n");
  for (int i = 1; i <= n; i++) printf("%d ", p[i]);
  printf("\n");
  for (int i = 1; i <= n; i++) printf("%d ", q[i]);
  printf("\n");
  return 0;
}