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>
const int maxn = 1e6 + 9, mod = 998244353, inf = 0x3f3f3f3f;
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 << 3ll) + (x << 1ll) + c - '0';
    c = getchar();
  }
  return x * f;
}
void Chkmin(int &x, int y) {
  if (y < x) x = y;
}
void Chkmax(int &x, int y) {
  if (y > x) x = y;
}
int add(int x, int y) { return x += y, x >= mod ? x - mod : x; }
int dec(int x, int y) { return x -= y, x < 0 ? x + mod : x; }
int mul(int x, int y) { return 1ll * x * y % mod; }
int Pow(int base, int b) {
  int ret(1);
  while (b) {
    if (b & 1) ret = mul(ret, base);
    base = mul(base, base);
    b >>= 1;
  }
  return ret;
}
int K;
int a[1 << 12], b[1 << 12], q[1 << 12], p[1 << 12], pos[1 << 12];
void Fix(int i, int j) {
  int k(pos[a[i] ^ q[i]]);
  if (k == i) return;
  if (k == j) {
    std::swap(p[i], p[j]);
    pos[p[i]] = i;
    pos[p[j]] = j;
    return;
  }
  std::swap(p[k], p[i]);
  pos[p[i]] = i;
  pos[p[k]] = k;
  std::swap(q[k], q[j]);
  Fix(k, j);
}
int main() {
  K = Read();
  int ret(0);
  for (int i = 0; i < (1 << K); ++i) {
    b[i] = Read();
    ret ^= b[i];
  }
  if (ret) {
    puts("Fou");
    return 0;
  }
  for (int i = 0; i < (1 << K); ++i) {
    p[i] = i;
    q[i] = i;
    a[i] = 0;
    pos[i] = i;
  }
  for (int i = 0; i < (1 << K); ++i)
    if (a[i] != b[i]) {
      int d(a[i] ^ b[i]);
      a[i] = b[i];
      a[i + 1] ^= d;
      Fix(i, i + 1);
    }
  for (int i = 0; i < (1 << K); ++i) {
    assert((p[i] ^ q[i]) == b[i]);
  }
  puts("Shi");
  for (int i = 0; i < (1 << K); ++i) {
    printf("%d ", p[i]);
  }
  puts("");
  for (int i = 0; i < (1 << K); ++i) {
    printf("%d ", q[i]);
  }
  puts("");
  return 0;
}