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>
#pragma optimize("SEX_ON_THE_BEACH")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC optimize("fast-math")
using ll = long long int;
using ull = long long int;
using dd = double;
using ldd = long double;
namespace someUsefull {
template <typename T1, typename T2>
inline void checkMin(T1& a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void checkMax(T1& a, T2 b) {
  if (a < b) a = b;
}
const int _popcount_size_ = 16;
char _bits_count_[1 << _popcount_size_];
inline void _build_bits_count_() {
  for (int i = 1, j = 0; i < (1 << _popcount_size_); ++i) {
    if (i & (1 << (j + 1))) ++j;
    _bits_count_[i] = _bits_count_[i ^ (1 << j)] + 1;
  }
}
template <typename T1 = int, typename T2 = int>
inline T1 popcnt(T2 x) {
  T1 ans = 0;
  while (x) {
    ans += _bits_count_[x & (1 << _popcount_size_) - 1];
    x >>= _popcount_size_;
  }
  return ans;
}
}  // namespace someUsefull
using namespace std;
using namespace someUsefull;
vector<int> pr;
vector<int> was;
void primes(int n) {
  was.resize(n + 1, 1e9);
  for (int i = 2; i < was.size(); ++i) {
    if (was[i] == int(1e9)) {
      pr.push_back(i);
    } else {
      continue;
    }
    for (int j = i; j < was.size(); j += i) checkMin(was[j], i);
  }
}
void solve(int test) {
  int n;
  cin >> n;
  if (n == 1) {
    cout << "C 1";
    return;
  }
  primes(n);
  vector<vector<int>> mp(n + 1);
  for (int i = 1; i <= n; ++i) {
    int j = i;
    while (j != 1) {
      mp[i].push_back(was[j]);
      j /= was[j];
    }
  }
  map<int, int> hoba;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < mp[i].size(); ++j) {
      if (j && mp[i][j] == mp[i][j - 1]) continue;
      ++hoba[mp[i][j]];
    }
  }
  0;
  ;
  0;
  ;
  0;
  ;
  vector<int> ans;
  for (int i = 0; i < pr.size(); ++i) {
    if ((ll)(pr[i]) * pr[i] > n) break;
    int good = hoba[pr[i]];
    cout << "B " << pr[i] << endl;
    int cnt;
    cin >> cnt;
    if (cnt != good) {
      ans.push_back(pr[i]);
    }
    for (int j = pr[i]; j <= n; j += pr[i]) {
      if (was[j] == pr[i]) {
        for (int k = 0; k < mp[j].size(); ++k) {
          if (k && mp[j][k] == mp[j][k - 1]) continue;
          hoba[mp[j][k]]--;
        }
      }
    }
  }
  int s = 1;
  for (ll j : pr) {
    if (j * j > n) ++s;
  }
  cout << "A 1" << endl;
  int k_;
  cin >> k_;
  if (s != k_) {
    for (int i = 0; i < pr.size(); ++i) {
      if ((ll)(pr[i]) * pr[i] <= n) continue;
      int good = hoba[pr[i]];
      cout << "B " << pr[i] << endl;
      int cnt;
      cin >> cnt;
      if (cnt != good) {
        ans.push_back(pr[i]);
      }
      for (int j = pr[i]; j <= n; j += pr[i]) {
        if (was[j] == pr[i]) {
          for (int k = 0; k < mp[j].size(); ++k) {
            if (k && mp[j][k] == mp[j][k - 1]) continue;
            hoba[mp[j][k]]--;
          }
        }
      }
    }
    0;
    ;
    ll x = 1;
    for (int i : ans) {
      x *= i;
      ll t = i;
      while (1) {
        t *= i;
        if (t > n) break;
        cout << "A " << t << endl;
        int l;
        cin >> l;
        if (l) {
          x *= i;
        } else {
          break;
        }
      }
    }
    for (int i = 0; i < pr.size(); ++i) {
      cout << "A " << pr[i] << endl;
      int c;
      cin >> c;
      if (c == 1) {
        ll t = pr[i];
        x *= pr[i];
        while (1) {
          t *= pr[i];
          if (t > n) break;
          cout << "A " << t << endl;
          int kk;
          cin >> kk;
          if (kk) {
            x *= pr[i];
          } else {
            break;
          }
        }
        break;
      }
    }
    cout << "C " << x;
    return;
  }
  reverse((pr).begin(), (pr).end());
  while (pr.size() && (ll)pr.back() * pr.back() <= n) pr.pop_back();
  reverse((pr).begin(), (pr).end());
  int block = 400;
  for (int i = 0; i < pr.size(); i += block) {
    for (int j = i; j < min(i + block, int(pr.size())); ++j) {
      cout << "B " << pr[j] << endl;
      int k;
      cin >> k;
      --k_;
    }
    cout << "A 1" << endl;
    int t;
    cin >> t;
    if (t != k_) {
      for (int j = i; j < min(i + block, int(pr.size())); ++j) {
        cout << "A " << pr[j] << endl;
        int k;
        cin >> k;
        if (k) {
          cout << "C " << pr[j];
          return;
        }
      }
    }
  }
  cout << "C 1";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  _build_bits_count_();
  int t = 1;
  for (int i = 0; i < t; ++i) {
    solve(i + 1);
  }
  return 0;
}