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 P = 1000000007;
const int INF = 1e9;
const int N = 1e5 + 5;
const double pi = acos(-1);
mt19937 rng(time(0));
bool isauto = false;
int n, p[N], tmp[N];
bool isp[N];
unordered_set<int> numbers;
int answer;
unordered_set<int> ansset;
int ask(int x, char op) {
  int res;
  cout << op << " " << x << endl;
  if (isauto) {
    res = 0;
    for (auto y : ansset)
      if (y % x == 0) tmp[++res] = y;
    if (op == 'B')
      for (int i = (1); i <= (res); ++i)
        if (tmp[i] != answer) ansset.erase(tmp[i]);
  } else
    cin >> res;
  return res;
}
int count(int x, bool del) {
  int res = 0;
  for (auto y : numbers)
    if (y % x == 0) tmp[++res] = y;
  if (del)
    for (int i = (1); i <= (res); ++i) numbers.erase(tmp[i]);
  return res;
}
int main() {
  cin >> n;
  if (isauto) answer = 99991;
  for (int i = (1); i <= (n); ++i) numbers.insert(i), ansset.insert(i);
  for (int i = (2); i <= (n); ++i) isp[i] = true;
  for (int i = (2); i <= (n); ++i)
    if (isp[i]) {
      for (int j = i + i; j <= n; j += i) isp[j] = false;
    }
  int nump = 0;
  for (int i = (2); i <= (n); ++i)
    if (isp[i]) p[++nump] = i;
  int numa = 0, interval = sqrt(nump);
  int l = -1, r = -1;
  vector<int> factors;
  for (int i = (1); i <= (nump); ++i) {
    int num;
    num = ask(p[i], 'B');
    if (num != count(p[i], true)) factors.push_back(p[i]);
    if (i % interval == 0) {
      num = ask(1, 'A');
      if (num != count(1, false) && l == -1) l = i - interval + 1, r = i;
    }
  }
  if (nump % interval) {
    int num = ask(1, 'A');
    if (num != count(1, false) && l == -1)
      l = nump / interval * interval + 1, r = nump;
  }
  if (l == -1) {
    cout << "C 1\n";
    return 0;
  }
  for (int i = (l); i <= (r); ++i)
    if (ask(p[i], 'A') != count(p[i], false)) {
      factors.push_back(p[i]);
      break;
    }
  int res = 1;
  for (auto x : factors) res *= x;
  for (auto x : factors) {
    long long tmp = 1ll * res * x;
    while (tmp <= n && ask(tmp, 'A')) {
      res = tmp;
      tmp *= x;
    }
  }
  cout << "C " << res << endl;
  return 0;
}
