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;
const int sz = 1e5 + 1;
int n;
int s[sz];
vector<int> prime;
void sieve() {
  for (int i = 2; i <= n; i++) {
    if (s[i] == 0) {
      prime.push_back(i);
      for (int j = i; j <= n; j += i) s[j] = i;
    }
  }
}
int ask(int type, int x) {
  if (type == 0)
    cout << "A " << x << endl;
  else if (type == 1)
    cout << "B " << x << endl;
  else {
    cout << "C " << x << endl;
    return 0;
  }
  int res;
  cin >> res;
  return res;
}
void solve() {
  cin >> n;
  sieve();
  fill(s, s + sz, 0);
  bool gff = 0;
  int l = 0, r = 0;
  vector<int> fact;
  int c = 0;
  int cur_st = n;
  int sq = sqrt(n) + 1;
  for (int x : prime) {
    int res = ask(1, x);
    int cnt1 = 0, cnt2 = 0;
    for (int i = x; i <= n; i += x) {
      if (s[i] != 0 && s[i] < x) cnt1++;
      if (s[i] == 0) cur_st--;
      cnt2++;
      s[i] = x;
    }
    if (res > cnt2 - cnt1) {
      fact.push_back(x);
      if (!gff) {
        gff = 1;
        r = c - 1;
      }
    }
    if (!gff && (c - l >= sq || c == 0)) {
      res = ask(0, 1);
      if (cur_st < res) {
        r = c;
        gff = 1;
      } else {
        l = c;
      }
    }
    c++;
  }
  if (!gff) {
    r = c - 1;
    gff = 1;
  }
  int res = ask(0, 1);
  if (res == 1) {
    ask(2, 1);
    return;
  }
  for (int i = l; i <= r; i++) {
    int res = ask(0, prime[i]);
    if (res == 1) {
      fact.push_back(prime[i]);
      break;
    }
  }
  int ans = 1;
  for (int x : fact) {
    long long cur = (long long)x * x;
    while (cur <= n && ask(0, cur) == 1) cur *= x;
    cur /= x;
    ans *= cur;
  }
  ask(2, ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  for (; q > 0; q--) {
    solve();
  }
}
