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("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
using namespace std;
namespace _c {
const double pi = acos(-1.0);
namespace min {
const int i8 = -128;
const int i16 = -32768;
const int i = -2147483647 - 1;
const long long l = -9223372036854775807LL - 1;
}  // namespace min
namespace max {
const int i8 = 127;
const int i16 = 32767;
const int i = 2147483647;
const long long l = 9223372036854775807LL;
}  // namespace max
}  // namespace _c
namespace _f {
template <typename T>
inline const T gcd(T m, T n) {
  while (n != 0) {
    T t = m % n;
    m = n;
    n = t;
  }
  return m;
}
template <typename T>
inline const T max(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T>
inline const T min(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline const T abs(const T &a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T pow(T a, T b) {
  T res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b >>= 1;
  }
  return res;
}
template <typename T>
inline T pow(T a, T b, const T &m) {
  a %= m;
  T res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % m;
    }
    a = a * a % m;
    b >>= 1;
  }
  return res % m;
}
}  // namespace _f
namespace io {
template <typename T>
inline void read(T &t) {
  register T res = 0, neg = 1;
  char g = getchar();
  for (; !isdigit(g); g = getchar()) {
    if (g == '-') {
      neg = -1;
    }
  }
  for (; isdigit(g); g = getchar()) {
    res = res * 10 + g - '0';
  }
  t = res * neg;
}
inline int in() {
  register int T;
  read(T);
  return T;
}
inline long long in_ll() {
  register long long T;
  read(T);
  return T;
}
}  // namespace io
int K, N;
vector<int> A, B, C;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> K;
  N = 1 << K;
  A.resize(N);
  int all_xor = 0;
  for (auto &a : A) {
    cin >> a;
    all_xor ^= a;
  }
  if (all_xor != 0) {
    cout << "Fou" << '\n';
    return 0;
  }
  B.resize(N);
  C.resize(N);
  vector<int> location(N);
  for (int i = 0; i < N; i++) {
    B[i] = C[i] = i;
    location[C[i]] = i;
  }
  for (int i = 0; i < N; i++) {
    int u = i;
    while ((A[u] ^ B[u] ^ C[u]) != 0) {
      int v = location[A[u] ^ B[u]];
      swap(C[u], C[v]);
      swap(location[C[u]], location[C[v]]);
      if (v > i) {
        break;
      } else {
        swap(B[v], B[i + 1]);
        u = v;
      }
    }
  }
  cout << "Shi" << '\n';
  for (int i = 0; i < N; i++) cout << B[i] << (i < N - 1 ? ' ' : '\n');
  for (int i = 0; i < N; i++) cout << C[i] << (i < N - 1 ? ' ' : '\n');
}