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 oo = 1e9 + 5;
const long long int ooll = (long long int)1e18 + 5;
const int MOD = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
clock_t start = clock();
const int N = (int)1e5 + 1;
const int BLOCK = 98;
int v[N], sp[N], pres[N], n, tot, gone[N];
bool debug = false;
int judge_a[N], x, MX_QUERIES = 9999, Q = 0;
int ask(int a, char type) {
  ++Q;
  assert(Q <= MX_QUERIES);
  if (debug) {
    int got = 0;
    for (int j = a; j <= n; j += a) {
      if (judge_a[j] == 1) ++got;
      if (type == 'B' && j != x) judge_a[j] = 0;
    }
    return got;
  }
  cout << type << " " << a << '\n';
  fflush(stdout);
  if (type == 'B') {
    for (int j = a; j <= n; j += a) {
      if (gone[j] == 1) {
        tot--;
      }
      gone[j] = 0;
    }
  }
  int ans;
  cin >> ans;
  return ans;
}
void pre() {
  for (int i = 1; i <= N; ++i) sp[i] = i;
  for (int i = 2; i < N; ++i)
    if (sp[i] == i) {
      if (i * 1LL * i >= N) break;
      for (int j = i + i; j < N; j += i) sp[j] = min(sp[j], i);
    }
}
void solve() {
  vector<int> primes;
  cin >> n;
  Q = 0;
  if (debug) cin >> x;
  tot = n;
  for (int i = 0; i <= n; ++i) {
    pres[i] = 0;
    judge_a[i] = 1;
    gone[i] = 1;
  }
  for (int i = 2; i <= n; ++i) {
    if (sp[i] == i) primes.push_back(i);
    pres[sp[i]]++;
  }
  vector<int> have;
  int psz = primes.size(), sj = psz - 1, nb = (psz - 1) % BLOCK;
  bool found = false;
  for (int j = 0; j < psz; ++j) {
    int it = primes[j];
    int go = ask(it, 'B');
    if (go != pres[it]) {
      have.push_back(it);
    }
    if (j % BLOCK == nb && !found) {
      int go = ask(1, 'A');
      if (go != tot) {
        sj = j;
        found = true;
      }
    }
  }
  if (!have.empty()) {
    for (int j = 0; j < psz; ++j) {
      int it = primes[j];
      if (it * 1LL * it >= n) break;
      int go = ask(it, 'A');
      if (go > 0) {
        have.push_back(it);
        break;
      }
    }
    int ans = 1;
    for (auto it : have) {
      int pw = it;
      assert(it > 1);
      while (pw * 1LL * it <= n) {
        pw *= it;
        int go = ask(pw, 'A');
        if (go == 0) {
          pw /= it;
          break;
        }
      }
      ans *= pw;
    }
    if (debug) {
      assert(x == ans);
    } else {
      cout << "C " << ans << '\n';
      fflush(stdout);
    }
    return;
  }
  int l = max(0, sj - BLOCK + 1), r = sj;
  for (int j = l; j <= r; ++j) {
    int go = ask(primes[j], 'A');
    if (go > 0) {
      int pw = primes[j];
      while (pw * 1LL * primes[j] <= n) {
        pw *= primes[j];
        int go = ask(pw, 'A');
        if (go == 0) {
          pw /= primes[j];
          break;
        }
      }
      int ans = pw;
      if (debug) {
        assert(x == ans);
      } else {
        cout << "C " << ans << '\n';
        fflush(stdout);
      }
      return;
    }
  }
  if (debug) {
    assert(x == 1);
  } else {
    cout << "C 1\n";
    fflush(stdout);
  }
}
int main() {
  int T = 1;
  pre();
  for (int t = 1; t <= T; ++t) {
    solve();
  }
  return 0;
}
