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>
#pragma GCC optimize(3)
using namespace std;
inline long long read() {
  long long x = 0, y = 0, c = getchar();
  while (!isdigit(c)) y = c, c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ '0'), c = getchar();
  return y == '-' ? -x : x;
}
inline void print(long long q) {
  if (q < 0) putchar('-'), q = -q;
  if (q >= 10) print(q / 10);
  putchar(q % 10 + '0');
}
long long s1 = 19260817, s2 = 23333333, s3 = 998244353, srd;
inline long long rd() { return srd = (srd * s1 + s2 + rand()) % s3; }
void file() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
int n, m, a[4100], p[4100], q[4100], b[4100], wz[4100];
inline void wk(int x, int y) {
  if ((b[x] ^ p[x]) == q[x]) return;
  int t = wz[b[x] ^ q[x]];
  if (t == y) {
    swap(p[x], p[y]);
    swap(wz[p[x]], wz[p[y]]);
    return;
  }
  swap(p[x], p[t]);
  swap(wz[p[x]], wz[p[t]]);
  swap(q[y], q[t]);
  wk(t, y);
}
int main() {
  srand(time(0));
  rd();
  m = read();
  n = 1 << m;
  int i, s = 0;
  for (i = 0; i < n; ++i) a[i] = read(), s ^= a[i];
  if (s) {
    puts("Fou");
    return 0;
  }
  puts("Shi");
  for (i = 0; i < n; ++i) p[i] = q[i] = wz[i] = i;
  for (i = 0; i < n - 1; ++i) {
    if (b[i] == a[i]) continue;
    b[i + 1] ^= b[i] ^ a[i];
    b[i] = a[i];
    wk(i, i + 1);
  }
  for (i = 0; i < n; ++i) print(p[i]), putchar(' ');
  putchar('\n');
  for (i = 0; i < n; ++i) print(q[i]), putchar(' ');
  putchar('\n');
  return 0;
}