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;
const long long N = 100100;
vector<long long> primes;
bool mark[N];
long long n;
void crivo() {
  for (long long i = 2; i <= n; i++) {
    if (!mark[i]) {
      if (i <= n) primes.push_back(i);
      for (long long j = 2 * i; j <= n; j += i) mark[j] = 1;
    }
  }
}
bool saiu[N];
long long askB(long long p) {
  if (p > n) return 0;
  cout << 'B' << " " << p << '\n';
  cout.flush();
  long long ans;
  cin >> ans;
  return ans;
}
long long askA(long long p) {
  if (p > n) return 0;
  cout << "A " << p << '\n';
  cout.flush();
  long long ans;
  cin >> ans;
  return ans;
}
long long get(long long p) {
  long long res = 0;
  for (long long i = p; i <= n; i += p) {
    if (saiu[i]) continue;
    res++;
    saiu[i] = 1;
  }
  return res;
}
long long getlog(long long x, long long mx) {
  long long r = 1;
  long long p = x;
  while (p * x <= mx) {
    p *= x;
    r++;
  }
  return r;
}
long long elevate(long long p, long long pwr) {
  long long r = 1;
  for (long long i = 0; i < pwr; i++) r *= p;
  return r;
}
const long long sq = 100;
long long P = 1;
void solve(vector<long long> cur) {
  for (long long p : cur) {
    long long x = askB(p);
    if (x != 0) {
      long long l = 2, r = getlog(p, n / P);
      long long res = 1;
      while (l <= r) {
        long long mid = (l + r) / 2;
        long long per = askA(P * elevate(p, mid));
        if (per != 0) {
          res = mid, l = mid + 1;
        } else
          r = mid - 1;
      }
      P *= elevate(p, res);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  crivo();
  vector<long long> cur;
  long long fora = 0;
  long long i = 0;
  for (i = 0; i <= (long long)(primes).size(); i++) {
    if (i == (long long)(primes).size()) {
      solve(cur);
      break;
    }
    cur.push_back(primes[i]);
    long long x = askB(primes[i]);
    long long s = get(primes[i]);
    fora += x;
    if (s != x) {
      solve(cur);
      break;
    }
    if ((long long)(cur).size() == sq) {
      long long A = askA(1);
      if (A != n - fora) {
        solve(cur);
        break;
      }
      cur.clear();
    }
  }
  if (P == 1) {
    cout << "C " << 1 << '\n';
    cout.flush();
    return 0;
  }
  i++;
  for (; i < (long long)(primes).size(); i++) {
    long long p = primes[i];
    if (p <= 1 || p > n) continue;
    long long per = askA(primes[i] * P);
    if (per != 0) {
      long long l = 2, r = getlog(p, n / P);
      long long res = 1;
      while (l <= r) {
        long long mid = (l + r) / 2;
        per = askA(elevate(p, mid) * P);
        if (per != 0) {
          l = mid + 1;
          res = mid;
        } else
          r = mid - 1;
      }
      P *= elevate(p, res);
    }
  }
  cout << "C " << P << '\n';
  cout.flush();
  return 0;
}