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;
const int N = 1e5 + 5;
int n, g_n;
int prime[N], tot;
int g_prime[N], cnt;
bool flag[N];
vector<pair<int, int> > g;
int L[N], R[N], id[N];
inline int min(int a, int b) { return a > b ? b : a; }
inline void pre() {
  for (register int i = 2; i <= n; ++i) {
    if (!flag[i]) {
      prime[++tot] = i;
    }
    for (register int j = 1; j <= tot; ++j) {
      if (i * prime[j] > n) break;
      flag[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
inline void get_pig(int pos) {
  for (register int i = L[pos]; i <= R[pos]; ++i) {
    int tot;
    cout << "A " << g_prime[i] << endl;
    cin >> tot;
    if (tot) {
      g.push_back(make_pair(g_prime[i], 1));
      return;
    }
  }
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    puts("C 1");
    fflush(stdout);
    return 0;
  }
  pre();
  g_n = sqrt(n);
  for (register int i = 1; i <= tot; ++i) {
    if (prime[i] <= g_n)
      g_prime[++cnt] = prime[i];
    else
      break;
  }
  for (register int i = 1; i <= cnt; ++i) {
    int x = g_prime[i];
    cout << "B " << x << endl;
    int cnt;
    cin >> cnt;
    cout << "A " << x << endl;
    cin >> cnt;
    int k = 0, res = x;
    if (cnt) {
      while (cnt) {
        ++k;
        res *= x;
        if (res > n) break;
        cout << "A " << res << endl;
        cin >> cnt;
      }
      g.push_back(make_pair(x, k));
    }
  }
  for (register int i = cnt + 1; i <= tot; ++i) {
    g_prime[++prime[0]] = prime[i];
  }
  cnt = prime[0];
  int size = sqrt(cnt);
  int sum_g = 0;
  if (cnt % size)
    sum_g = cnt / size + 1;
  else
    sum_g = cnt / size;
  for (register int i = 1; i <= sum_g; ++i) {
    L[i] = (i - 1) * size + 1;
    R[i] = min(i * size, cnt);
  }
  for (register int i = 1; i <= sum_g; ++i)
    for (register int j = L[i]; j <= R[i]; ++j) id[j] = i;
  cout << "A 1" << endl;
  cin >> cnt;
  for (register int k = 1; k <= sum_g; ++k) {
    int expect = 0;
    for (register int i = L[k]; i <= R[k]; ++i) {
      int x;
      cout << "B " << g_prime[i] << endl;
      cin >> x;
      expect += x;
    }
    int tot;
    cout << "A 1" << endl;
    cin >> tot;
    if (cnt - tot != expect) {
      get_pig(k);
      break;
    }
    cnt = tot;
  }
  int ans = 1;
  int sx = g.size();
  for (register int k = 0; k < sx; ++k) {
    int x = g[k].first, p = g[k].second;
    while (p) {
      ans *= x;
      --p;
    }
  }
  cout << "C " << ans << endl;
  return 0;
}