Problem:
Score : 100 points

Problem Statement

Decades have passed since the beginning of AtCoder Beginner Contest.

The contests are labeled as ABC001, ABC002, ... from the first round, but after the 999-th round ABC999, a problem occurred: how the future rounds should be labeled?

In the end, the labels for the rounds from the 1000-th to the 1998-th are decided: ABD001, ABD002, ..., ABD999.

You are given an integer N between 1 and 1998 (inclusive). Print the first three characters of the label of the N-th round of AtCoder Beginner Contest.

Constraints

1 \leq N \leq 1998

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the first three characters of the label of the N-th round of AtCoder Beginner Contest.

Sample Input 1

999

Sample Output 1

ABC

The 999-th round of AtCoder Beginner Contest is labeled as ABC999.

Sample Input 2

1000

Sample Output 2

ABD

The 1000-th round of AtCoder Beginner Contest is labeled as ABD001.

Sample Input 3

1481

Sample Output 3

ABD

The 1481-th round of AtCoder Beginner Contest is labeled as ABD482.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <regex>
#include <numeric>
using namespace std;

/*--- Macro ---*/
#define rep(i, n) for(int i=0; i<(int)n; i++)
#define rep1(i, n) for(int i=1; i<(int)n; i++)
#define mp make_pair
#define pb push_back
#define all(a) a.begin(), a.end()

/*---  Type ---*/
typedef long long ll;

/*--- Constants ---*/
template <typename T> const T INF = numeric_limits<T>::max() / 10;
const int MOD = (int)1e9 + 7;

// Floating
const double EPS = 1e-11;
#define EQ0(x) (abs(x) < EPS)
#define EQ(x, y) (abs(a-b) < EPS)

// Util
int max(int a, int b, int c) {return max(a, max(b, c));}
int min(int a, int b, int c) {return min(a, min(b, c));}
ll max(ll x, ll y) {return x > y ? x : y;}
ll min(ll x, ll y) {return x < y ? x : y;}
ll max(ll a, ll b, ll c) {return max(a, max(b, c));}
ll min(ll a, ll b, ll c) {return min(a, min(b, c));}

// -------------------------------------------------------------------------------------



int main() {
  int n;
  cin >> n;

  if(n < 1000) cout << "ABC" << endl;
  else cout << "ABD" << endl;

  return 0;
}