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 MOD = 1e9 + 7;
const int maxn = 1e5 + 100;
const long long linf = 9223372036854775807;
const int inf = 1 << 31;
const double eps = 1e-4;
const double DoubleInf = 1e20;
const double pi = acos(-1.0);
long long qpow(long long a, long long b) {
  return b ? ((b & 1) ? a * qpow(a * a % MOD, b >> 1) % MOD
                      : qpow(a * a % MOD, b >> 1)) %
                 MOD
           : 1ll;
}
long long quick_mod(long long a, long long b, long long MOD) {
  long long ans = 1;
  a %= MOD;
  while (b) {
    if (b & 1) {
      ans = ans * a % MOD;
      b--;
    }
    b >>= 1;
    a = a * a % MOD;
  }
  return ans;
}
inline int ird() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x * 2) + ch - '0', ch = getchar();
  return x * f;
}
inline long long lrd() {
  register long long x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x * 2) + ch - '0', ch = getchar();
  return x * f;
}
void puts1() { cout << "-1" << endl; }
void puts2() { cout << "YES" << endl; }
void puts3() { cout << "NO" << endl; }
int gcd(int a, int b) { return (b > 0) ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
long long lgcd(long long a, long long b) {
  return (b > 0) ? lgcd(b, a % b) : a;
}
long long llcm(long long a, long long b) { return a * b / lgcd(a, b); }
int v[maxn], prime[maxn];
int primes(int n) {
  int m = 0;
  for (int i = 2; i <= n; i++) {
    if (v[i] == 0) {
      v[i] = i;
      prime[++m] = i;
    }
    for (int j = 1; j <= m; j++) {
      if (prime[j] > v[i] || prime[j] > n / i) break;
      v[i * prime[j]] = prime[j];
    }
  }
  return m;
}
int main() {
  int n, m, ans, M, g, start;
  int t;
  int tt, tot, temp, cur;
  tot = 0;
  cur = 0;
  g = 0;
  start = 0;
  M = primes(maxn);
  ans = 1;
  scanf("%d", &n);
  if (n == 1) {
    printf("C 1\n");
    return 0;
  }
  for (int i = 1; prime[i] <= n; i++) {
    cur++;
    if ((long long)(prime[i]) * (long long)(prime[i]) > (long long)(n))
      g++;
    else
      start = i;
  }
  M = cur;
  for (int i = 1; prime[i] * prime[i] <= n; i++) {
    cur--;
    printf("B %d\n", prime[i]);
    fflush(stdout);
    scanf("%d", &tt);
    printf("A %d\n", prime[i]);
    fflush(stdout);
    scanf("%d", &tt);
    if (tt) {
      ans *= prime[i];
      temp = prime[i];
      while (temp * prime[i] <= n) {
        temp *= prime[i];
        printf("A %d\n", temp);
        fflush(stdout);
        scanf("%d", &tt);
        if (tt == 0)
          break;
        else
          ans *= prime[i];
      }
    }
  }
  int s = (int)(sqrt(g));
  int pos;
  if (ans == 1) {
    for (int i = 0; i <= g / s - (g % s == 0); i++) {
      if (ans != 1) break;
      for (int j = 0; j < s; j++) {
        pos = start + i * s + j + 1;
        if (pos > M) continue;
        printf("B %d\n", prime[pos]);
        fflush(stdout);
        scanf("%d", &tt);
        cur--;
      }
      printf("A 1\n");
      fflush(stdout);
      scanf("%d", &tt);
      if (tt != cur + 1) {
        for (int j = 0; j < s; j++) {
          pos = start + i * s + j + 1;
          if (pos > M) continue;
          printf("A %d\n", prime[pos]);
          fflush(stdout);
          scanf("%d", &tt);
          if (tt) {
            ans *= prime[pos];
          }
        }
      }
    }
  } else {
    for (int i = start + 1; prime[i] <= n; i++) {
      printf("A %d\n", prime[i]);
      fflush(stdout);
      scanf("%d", &tt);
      if (tt == 2) {
        ans *= prime[i];
        break;
      }
    }
  }
  printf("C %d\n", ans);
  return 0;
}
