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;
struct IO {
  IO() {
    ios::sync_with_stdio(0);
    cin.tie(0);
  }
} io;
using ll = long long;
using db = long double;
using pi = pair<ll, ll>;
using mi = map<ll, ll>;
using vi = vector<ll>;
const ll N = 1e5 + 1, M = 1e9 + 7;
inline ll add(ll x, ll y) { return x + y >= M ? x + y - M : x + y; }
inline ll mul(ll x, ll y) { return x * y % M; }
inline ll sub(ll x, ll y) { return x - y < 0 ? x - y + M : x - y; }
ll pri[N], isp[N];
void init() {
  for (ll i = 2; i < N; i++) {
    if (!isp[i]) {
      pri[++*pri] = i;
    }
    for (ll j = 1; j <= *pri && pri[j] * i < N; j++) {
      isp[pri[j] * i] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
signed main() {
  init();
  ll n;
  cin >> n;
  ll ans = 1;
  for (ll i = 1; i < *pri; i++) {
    if (pri[i] * pri[i] <= n) {
      cout << "B"
           << " " << pri[i] << endl;
      cout.flush();
      ll res;
      cin >> res;
      cout << "A"
           << " " << pri[i] << endl;
      cout.flush();
      cin >> res;
      if (res) {
        ans *= pri[i];
        ll g = pri[i] * pri[i];
        while (g <= n) {
          cout << "A"
               << " " << g << endl;
          cout.flush();
          cin >> res;
          g *= pri[i];
          if (res) {
            ans *= pri[i];
          } else
            break;
        }
      }
    }
  }
  if (ans > 1) {
    for (ll i = 1; i <= *pri; i++) {
      if (pri[i] * pri[i] > n && pri[i] <= n / ans) {
        ll res;
        cout << "A"
             << " " << pri[i] << endl;
        cout.flush();
        cin >> res;
        if (res == 2) {
          ans *= pri[i];
          break;
        }
      }
    }
  } else {
    for (ll i = *pri; i >= 1; i--)
      if (pri[i] > n) --*pri;
    ll l = 1, r = *pri;
    while (l <= r) {
      ll mid = l + r >> 1;
      ll res;
      for (ll i = l; i <= mid; i++) {
        cout << "B " << pri[i] << endl;
        cin >> res;
      }
      cout << "A " << 1 << endl;
      cin >> res;
      if (res == 1 + r - mid) {
        l = mid + 1;
      } else {
        for (ll i = l; i <= mid; i++) {
          cout << "A " << pri[i] << endl;
          cin >> res;
          if (res) {
            ans = pri[i];
            break;
          }
        }
        break;
      }
    }
  }
  cout << "C " << ans << endl;
}
