Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct SIEVE_OF_ERATOSTHENES {
  vector<long long> primes;
  vector<long long> least;
  vector<long long> cnt;
  vector<long long> pw;
  vector<long long> phi;
  vector<long long> sigma;
  vector<long long> mu;
  SIEVE_OF_ERATOSTHENES(long long N) {
    if (N < 1) N = 1;
    least.resize(N + 1);
    cnt.resize(N + 1);
    pw.resize(N + 1);
    phi.resize(N + 1);
    sigma.resize(N + 1);
    mu.resize(N + 1);
    pw[1] = 0;
    least[1] = cnt[1] = phi[1] = sigma[1] = mu[1] = 1;
    for (long long i = 4; i <= N; i += 2) least[i] = 2;
    for (long long i = 3; i * i <= N; i += 2) {
      if (least[i] == 0) {
        for (long long j = i * i; j <= N; j += i * 2) {
          if (least[j] == 0) least[j] = i;
        }
      }
    }
    for (long long i = 2; i <= N; i++) {
      if (least[i] == 0) {
        primes.push_back(i);
        least[i] = i;
        cnt[i] = i;
        pw[i] = 1;
        phi[i] = i - 1;
        sigma[i] = i + 1;
        mu[i] = -1;
      } else {
        long long x = i / least[i];
        if (least[i] == least[x])
          cnt[i] = cnt[x] * least[i], pw[i] = pw[x] + 1;
        else
          cnt[i] = least[i], pw[i] = 1;
        if (i == cnt[i]) {
          phi[i] = cnt[i] - cnt[i] / least[i];
          sigma[i] = (cnt[i] * least[i] - 1) / (least[i] - 1);
          mu[i] = 0;
        } else {
          phi[i] = phi[i / cnt[i]] * phi[cnt[i]];
          sigma[i] = sigma[i / cnt[i]] * sigma[cnt[i]];
          mu[i] = mu[i / cnt[i]] * mu[cnt[i]];
        }
      }
    }
  }
};
long long N;
long long AAA(long long x) {
  if (x > N) return 0;
  cout << "A"
       << " " << x << '\n';
  cout.flush();
  long long qu;
  cin >> qu;
  return qu;
}
long long BBB(long long x) {
  if (x > N) return 0;
  cout << "B"
       << " " << x << '\n';
  cout.flush();
  long long qu;
  cin >> qu;
  return qu;
}
void CCC(long long x) {
  cout << "C"
       << " " << x << '\n';
  cout.flush();
}
int32_t main() {
  cin >> N;
  SIEVE_OF_ERATOSTHENES CRIBA(N);
  vector<long long> primos = CRIBA.primes;
  vector<long long> fact;
  vector<long long> temp;
  for (long long i = 0; i < primos.size(); i++) {
    if (primos[i] * primos[i] > N) {
      temp.push_back(primos[i]);
      continue;
    }
    long long qu1 = BBB(primos[i]);
    long long qu2 = AAA(primos[i]);
    if (qu2 == 1) {
      fact.push_back(primos[i]);
    }
  }
  primos = temp;
  const long long BLOCK = 100;
  vector<vector<long long>> arr;
  vector<long long> ttt;
  for (auto x : primos) {
    ttt.push_back(x);
    if (ttt.size() == BLOCK) {
      arr.push_back(ttt);
      ttt.clear();
    }
  }
  if (ttt.size()) arr.push_back(ttt);
  long long total = AAA(1);
  long long temporal;
  if (fact.size())
    temporal = 1;
  else
    temporal = 0;
  for (auto vect : arr) {
    long long sz = vect.size();
    for (auto x : vect) {
      long long quu = BBB(x);
      if (temporal == 1 && quu == 2) fact.push_back(x);
    }
    long long tot = AAA(1);
    if (total - tot != sz) {
      for (auto x : vect) {
        if (AAA(x) == 1) fact.push_back(x);
      }
      break;
    }
    total = tot;
  }
  long long num = 1;
  for (auto x : fact) {
    long long tt = x;
    while (1) {
      if (AAA(tt) == 1) {
        num *= x;
        tt *= x;
      } else {
        break;
      }
    }
  }
  CCC(num);
  return 0;
}