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;
constexpr int seed = 131;
constexpr long long LLMAX = 2e18;
constexpr int MOD = 1e9 + 7;
constexpr double eps = 1e-8;
constexpr int MAXN = 1e6 + 10;
constexpr int hmod1 = 0x48E2DCE7;
constexpr int hmod2 = 0x60000005;
inline long long sqr(long long x) { return x * x; }
inline int sqr(int x) { return x * x; }
inline double sqr(double x) { return x * x; }
inline int mul(int x, int y) { return 1ll * x * y % MOD; }
inline int sub(int x, int y) { return x >= y ? x - y : MOD + x - y; }
inline int add(int x, int y) { return x + y >= MOD ? x + y - MOD : x + y; }
inline int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return (x > 0 ? 1 : -1);
}
inline long long qpow(long long a, long long n) {
  long long sum = 1;
  while (n) {
    if (n & 1) sum = sum * a % MOD;
    a = a * a % MOD;
    n >>= 1;
  }
  return sum;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long exgcd(long long a, long long b, long long &x, long long &y) {
  long long d;
  (b == 0 ? (x = 1, y = 0, d = a)
          : (d = exgcd(b, a % b, y, x), y -= a / b * x));
  return d;
}
vector<pair<long long, long long> > fac;
bool is[MAXN];
int a[MAXN], prime[MAXN], cnt;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 2; i < MAXN; i++) {
    if (!a[i]) prime[cnt++] = i;
    for (int j = 0; j < cnt && i * prime[j] < MAXN; j++) {
      a[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  int i, n;
  cin >> n;
  int left = n;
  for (i = 0; prime[i] * prime[i] <= n; i++) {
    cout << "B " << prime[i] << endl;
    cout.flush();
    int x;
    cin >> x;
    cout << "A " << prime[i] << endl;
    cout.flush();
    int y;
    cin >> y;
    for (int j = prime[i]; j <= n; j += prime[i]) left -= !is[j], is[j] = 1;
    if (y) {
      fac.push_back({prime[i], 1});
      for (i++; prime[i] <= n; i++) {
        cout << "B " << prime[i] << endl;
        cout.flush();
        int x;
        cin >> x;
        for (int j = prime[i]; j <= n; j += prime[i]) x -= !is[j], is[j] = 1;
        if (x) fac.push_back({prime[i], 1});
      }
      break;
    }
  }
  if (fac.empty()) {
    for (int s = i; prime[i] <= n; i++) {
      cout << "B " << prime[i] << endl;
      cout.flush();
      int x;
      cin >> x;
      for (int j = prime[i]; j <= n; j += prime[i]) left -= !is[j], is[j] = 1;
      if (prime[i + 1] > n || i - s == 100) {
        cout << "A " << 1 << endl;
        cout.flush();
        int y;
        cin >> y;
        if (left != y) {
          for (int j = s; j <= i; j++) {
            cout << "A " << prime[j] << endl;
            cout.flush();
            int z;
            cin >> z;
            if (z) {
              fac.push_back({prime[j], 1});
              i = cnt - 2;
              break;
            }
          }
        }
        s = i + 1;
      }
    }
  }
  int ans = 1;
  for (int i = 0; i < (int)fac.size(); i++) {
    for (long long j = fac[i].first * fac[i].first; j <= n; j *= fac[i].first) {
      cout << "A " << j << endl;
      cout.flush();
      int x;
      cin >> x;
      if (x)
        fac[i].second++;
      else
        break;
    }
  }
  for (auto i : fac) ans *= qpow(i.first, i.second);
  cout << "C " << ans << endl;
  cout.flush();
  return 0;
}
