This is an interactive problem.

There is an unknown integer x (1≤ x≤ n). You want to find x.

At first, you have a set of integers \{1, 2, …, n\}. You can perform the following operations no more than 10000 times:

  * A a: find how many numbers are multiples of a in the current set. 
  * B a: find how many numbers are multiples of a in this set, and then delete all multiples of a, but x will never be deleted (even if it is a multiple of a). In this operation, a must be greater than 1. 
  * C a: it means that you know that x=a. This operation can be only performed once. 



Remember that in the operation of type B a>1 must hold.

Write a program, that will find the value of x.

Input

The first line contains one integer n (1≤ n≤ 10^5). The remaining parts of the input will be given throughout the interaction process.

Interaction

In each round, your program needs to print a line containing one uppercase letter A, B or C and an integer a (1≤ a≤ n for operations A and C, 2≤ a≤ n for operation B). This line desribes operation you make.

If your operation has type C your program should terminate immediately.

Else your program should read one line containing a single integer, which is the answer to your operation.

After outputting each line, don't forget to flush the output. To do it use:

  * fflush(stdout) in C/C++; 
  * System.out.flush() in Java; 
  * sys.stdout.flush() in Python; 
  * flush(output) in Pascal; 
  * See the documentation for other languages. 



It is guaranteed, that the number x is fixed and won't change during the interaction process.

Hacks:

To make a hack, use such input format:

The only line should contain two integers n, x (1 ≤ x ≤ n ≤ 10^5).

Example

Input


10

2

4

0

Output



B 4

A 2

A 8

C 4

Note

Note that to make the sample more clear, we added extra empty lines. You shouldn't print any extra empty lines during the interaction process.

In the first test n=10 and x=4.

Initially the set is: \{1,2,3,4,5,6,7,8,9,10\}.

In the first operation, you ask how many numbers are multiples of 4 and delete them. The answer is 2 because there are two numbers divisible by 4: \{4,8\}. 8 will be deleted but 4 won't, because the number x will never be deleted. Now the set is \{1,2,3,4,5,6,7,9,10\}.

In the second operation, you ask how many numbers are multiples of 2. The answer is 4 because there are four numbers divisible by 2: \{2,4,6,10\}.

In the third operation, you ask how many numbers are multiples of 8. The answer is 0 because there isn't any number divisible by 8 in the current set.

In the fourth operation, you know that x=4, which is the right answer.
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using ll = long long;
const int N = 1e5 + 1;
bool isPrime[N];
vector<int> primes;
int mnDiv[N];
void sieve() {
  for (int i = 2; i < N; i++) isPrime[i] = true, mnDiv[i] = -1;
  for (int i = 2; i < N; i++) {
    if (isPrime[i]) {
      primes.push_back(i);
      mnDiv[i] = i;
      for (int j = i + i; j < N; j += i) {
        if (mnDiv[j] == -1) mnDiv[j] = i;
        isPrime[j] = false;
      }
    }
  }
}
const int SQ = 100;
int ddd;
int ask(char c, int x) {
  if (ddd > 9990) {
    while (true)
      ;
    exit(0);
  }
  cout << c << " " << x << endl;
  fflush(stdout);
  ddd++;
  if (c == 'C') exit(0);
  int y;
  cin >> y;
  return y;
}
int n;
int cnt[N];
int main() {
  ios_base::sync_with_stdio(false);
  sieve();
  cin >> n;
  cnt[1] = n;
  for (int i = 2; i <= n; i++) cnt[mnDiv[i]]++;
  int d = 0;
  vector<int> factors;
  bool first = false;
  for (int i = 0; i < primes.size(); i++) {
    d++;
    if (d % SQ == 0 && !first) {
      int ans = ask('A', 1);
      if (ans != cnt[1]) {
        first = true;
        for (int j = d - SQ; j < d; j++) {
          int y = primes[j];
          int ans = ask('A', y);
          if (ans != 0) {
            factors.push_back(y);
            break;
          }
        }
      }
    }
    int x = primes[i];
    if (x > n) break;
    int ans = ask('B', x);
    cnt[1] -= cnt[x];
    if (ans != cnt[x]) factors.push_back(x);
    cnt[x] = 0;
  }
  if (!first) {
    int ans = ask('A', 1);
    if (ans != cnt[1]) {
      first = true;
      for (int j = max(0, d - SQ); j < primes.size(); j++) {
        int y = primes[j];
        if (y > n) break;
        int ans = ask('A', y);
        if (ans != 0) {
          factors.push_back(y);
          break;
        }
      }
    }
  }
  for (auto &x : factors) {
    ll cur = 1LL * x * x;
    while (cur <= n) {
      ll ans = ask('A', cur);
      if (ans == 0) break;
      cur *= x;
    }
    cur /= x;
    x = cur;
  }
  ll res = 1;
  for (auto x : factors) res *= x;
  ask('C', res);
  return 0;
}
