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 maxn = 1e5 + 5;
const int local = 0;
int ask = 0;
set<int> s;
int X, N;
int A(int num) {
  if (num > N) return 0;
  ++ask;
  if (!local) {
    cout << "A " << num << endl;
    int x;
    cin >> x;
    return x;
  } else {
    int ret = 0;
    for (int it : s) {
      if (it % num == 0) ++ret;
    }
    return ret;
  }
}
int B(int num) {
  if (num > N) return 0;
  ++ask;
  assert(num != 1);
  if (!local) {
    cout << "B " << num << endl;
    int x;
    cin >> x;
    return x;
  } else {
    int ret = 0;
    vector<int> del;
    for (int a : s) {
      if (a % num == 0) {
        ret++;
        if (a != X) del.push_back(a);
      }
    }
    for (int a : del) s.erase(a);
    return ret;
  }
}
void C(int num) {
  if (local) {
    assert(num == X);
  }
  ++ask;
  cout << "C " << num;
}
vector<int> pr, small, big;
int res = 1;
void init(void) {
  cin >> X;
  for (int i = 1; i <= N; ++i) {
    s.insert(i);
  }
}
signed main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  auto prime = [&](int x) {
    for (int i = 2; i * i <= x; ++i) {
      if (x % i == 0) {
        return false;
      }
    }
    return true;
  };
  cin >> N;
  if (local) {
    init();
  }
  for (int i = 2; i <= N; ++i) {
    if (prime(i)) {
      pr.emplace_back(i);
    }
  }
  int sq = sqrt(N);
  for (int& i : pr) {
    if (i <= sq) {
      B(i);
      small.emplace_back(i);
    } else {
      big.emplace_back(i);
    }
  }
  for (int& i : small) {
    int j = i;
    while (j <= N && A(j)) {
      j *= i;
    }
    j /= i;
    res *= j;
  }
  int num = A(1);
  bool ok = false;
  while (big.size()) {
    vector<int> v1, v2;
    for (int i = 0; i < (int)big.size(); ++i) {
      if (i & 1) {
        v2.emplace_back(big[i]);
      } else {
        v1.emplace_back(big[i]);
      }
    }
    int cnum = num;
    bool kt = false;
    for (int it : v1) {
      if (B(it) == 2) {
        res *= it;
        kt = true;
        break;
      }
      --cnum;
    }
    if (kt) break;
    num = A(1);
    if (cnum == num) {
      swap(big, v2);
    } else {
      swap(big, v1);
      ok = true;
      break;
    }
  }
  if (ok) {
    for (int& i : big) {
      if (B(i)) {
        res *= i;
      }
    }
  }
  cerr << ask;
  assert(ask <= 10000);
  C(res);
}
