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 long long maxn = 100010;
const long long mod = 1000000007;
const long double eps = 1e-9;
const long long inf = ((1ll << 31ll) - 1ll);
const long long INF = 10000000000000007ll;
const long double pi = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long qpow(long long b, long long e) {
  if (!e) return 1;
  if (e & 1) return qpow(b, e - 1) * b % mod;
  long long pwur = qpow(b, e >> 1);
  return pwur * pwur % mod;
}
long long modinv(long long x) { return qpow(x, mod - 2); }
long long n;
vector<long long> pr1, pr2;
vector<long long> block[505];
bool isp(long long x) {
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
long long qa(long long x) {
  cout << "A " << x << '\n';
  cout.flush();
  long long y;
  cin >> y;
  return y;
}
long long qb(long long x) {
  cout << "B " << x << '\n';
  cout.flush();
  long long y;
  cin >> y;
  return y;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(0);
  srand(time(NULL));
  cin >> n;
  if (n == 1) {
    cout << "C " << 1 << '\n';
    cout.flush();
    return 0;
  }
  if (n <= 3000) {
    long long ans = 1;
    for (long long i = 2; i <= n; i++)
      if (isp(i)) pr1.push_back(i);
    for (auto i : pr1) {
      long long x = qb(i);
      long long y = qa(i);
      if (y) {
        for (long long j = i; j <= n; j *= i) {
          long long z = qa(j);
          if (z == 0) {
            ans *= j / i;
            break;
          }
          if (j * i > n) {
            ans *= j;
          }
        }
      }
    }
    cout << "C " << ans << '\n';
    cout.flush();
    return 0;
  }
  long long ans = 1;
  for (long long i = 2; i <= sqrt(n); i++)
    if (isp(i)) pr1.push_back(i);
  for (long long i = sqrt(n) + 1; i <= n; i++)
    if (isp(i)) pr2.push_back(i);
  for (auto i : pr1) {
    long long x = qb(i);
    long long y = qa(i);
    if (y) {
      for (long long j = i * i; j <= n; j *= i) {
        long long z = qa(j);
        if (z == 0) {
          ans *= j / i;
          break;
        }
        if (j * i > n) {
          ans *= j;
        }
      }
    }
  }
  long long cnt = 0;
  long long sq = sqrt((long long)pr2.size());
  for (auto i : pr2) {
    block[cnt / sq].push_back(i);
    cnt++;
  }
  for (long long i = 0; i < 500; i++) {
    if (block[i].size() >= 1) {
      long long x = qa(1);
      long long sum = 0;
      for (auto j : block[i]) {
        sum += qb(j);
      }
      long long y = qa(1);
      if (x - (long long)block[i].size() == y &&
          sum == (long long)block[i].size())
        continue;
      for (auto j : block[i]) {
        if (qa(j)) ans *= j;
      }
    }
  }
  cout << "C " << ans << '\n';
  cout.flush();
  return 0;
}
