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;
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
int MOD = 1e9 + 7;
void ADD(long long &x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int SIZE = 1 << 20;
int p[SIZE];
vector<int> pp;
void prime_init() {
  for (int i = 2; i < SIZE; i++) {
    if (!p[i]) {
      pp.emplace_back(i);
      for (int j = i; j < SIZE; j += i) {
        if (!p[j]) p[j] = i;
      }
    }
  }
}
bool is_prime(int x) { return p[x] == x; }
bitset<SIZE> u;
int ask(char c, int x) {
  W(c, x);
  fflush(stdout);
  if (c == 'C') {
    exit(0);
  }
  int v;
  R(v);
  return v;
}
void solve() {
  prime_init();
  int n;
  R(n);
  vector<int> AA;
  for (int i = (2); i <= (n); ++i) {
    if (is_prime(i)) AA.emplace_back(i);
  }
  int sqn = sqrt(((int)(AA).size()) + 0.1);
  int rn = n;
  vector<int> ps;
  bool found = 0;
  for (int i = 0; i < ((int)(AA).size()); i += sqn) {
    int rr = min(((int)(AA).size()), i + sqn);
    for (int j = (i); j < (rr); ++j) {
      int real_v = ask('B', AA[j]);
      int expeted_v = 0;
      for (int k = AA[j]; k <= n; k += AA[j]) {
        if (!u[k]) {
          u[k] = 1;
          expeted_v++;
          rn--;
        }
      }
      if (real_v != expeted_v) {
        ps.emplace_back(AA[j]);
      }
    }
    if (!found) {
      int ret = ask('A', 1);
      if (ret != rn) {
        found = 1;
        for (int j = (i); j < (rr); ++j) {
          ret = ask('A', AA[j]);
          if (ret == 1) {
            ps.emplace_back(AA[j]);
            break;
          }
        }
      }
    }
  }
  long long an = 1;
  for (int i = 0; i < (((int)(ps).size())); ++i) {
    an *= ps[i];
    while (an * ps[i] <= n && ask('A', an * ps[i]) == 1) {
      an *= ps[i];
    }
  }
  ask('C', an);
}
int main() {
  solve();
  return 0;
}