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>
int vis[100005];
int small[100005], large[100005];
int delet[100005];
int size0 = 0, size1 = 0;
int l[100005], r[100005];
int remain[100005];
long long queue[105];
int keep[100005];
int tag[100005];
void sieve(int n) {
  for (int i = 1; i <= n; i++) vis[i] = 0;
  vis[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) {
      for (int j = 2 * i; j <= n; j += i) vis[j] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      small[++size0] = i;
    }
  }
}
int min(int a, int b) { return a > b ? b : a; }
void preWork(int n, int sq, int size) {
  remain[0] = n;
  for (int i = 1; (i - 1) * sq < size; i++) {
    remain[i] = remain[i - 1];
    l[i] = (i - 1) * sq + 1;
    r[i] = min(size, l[i] + sq - 1);
    for (int j = l[i]; j <= r[i]; j++) {
      int u = small[j];
      for (int k = u; k <= n; k += u) {
        if (!delet[k]) {
          delet[k] = 1;
          remain[i]--;
        }
      }
    }
  }
}
int nn = -1, mm = -1;
int ok = 1;
int doA(int u) {
  if (u < 1 || u > nn) {
    ok = 0;
    return -1;
  } else {
    int res = 0;
    for (int i = u; i <= nn; i += u) res += keep[i];
    return res;
  }
}
int doB(int u) {
  if (u < 1 || u > nn) {
    ok = 0;
    return -1;
  } else {
    int res = 0;
    for (int i = u; i <= nn; i += u) {
      res += keep[i];
      if (i != mm) keep[i] = 0;
    }
    return res;
  }
}
int all = 0;
int askA(int u) {
  all++;
  printf("A %d\n", u);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;
}
int askB(int u) {
  all++;
  printf("B %d\n", u);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;
}
void doDelet(int u, int n) {
  for (int i = u; i <= n; i += u) delet[i] = 1;
}
int getCnt(int u, int n) {
  int res = 0;
  for (int i = u; i <= n; i += u) res += (delet[i] == 0);
  return res;
}
int getAns(int size, long long n) {
  int res = 1;
  for (int i = 1; i <= n; i++) tag[i] = 0;
  for (int i = 1; i <= size; i++) {
    if (tag[queue[i]])
      continue;
    else
      tag[queue[i]] = 1;
    long long u = queue[i] * queue[i], mul = queue[i];
    while (u <= n && askA(u) != 0) {
      mul = u;
      u *= queue[i];
    }
    res = res * mul;
  }
  return res;
}
int getTest() {
  nn = rand() % 100 + 1;
  mm = rand() % nn + 1;
  for (int i = 1; i <= nn; i++) keep[i] = 1;
  return nn;
}
int main() {
  int n;
  scanf("%d", &n);
  all = 0;
  all = 0, ok = 1, size0 = 0;
  sieve(n);
  int sq = 100;
  for (int i = 1; i <= n; i++) delet[i] = 0;
  preWork(n, sq, size0);
  for (int i = 1; i <= n; i++) delet[i] = 0;
  int fst = -1, size = 0;
  for (int i = 1; (i - 1) * sq < size0; i++) {
    for (int j = l[i]; j <= r[i]; j++) {
      int u = small[j];
      int org = getCnt(u, n);
      int cur = askB(u);
      doDelet(u, n);
      if (cur != org) queue[++size] = u;
    }
    if (fst == -1) {
      int cur = askA(1);
      if (cur != remain[i]) {
        int find = -1;
        for (int j = l[i]; j <= r[i]; j++) {
          int u = small[j];
          if (askB(u) != 0) {
            find = u;
            break;
          }
        }
        fst = find;
      }
    }
  }
  if (fst != -1) queue[++size] = fst;
  int res = getAns(size, n);
  printf("C %d\n", res);
  return 0;
}