Problem:
Score : 100 points

Problem Statement

In AtCoder City, there are three stations numbered 1, 2, and 3.

Each of these stations is operated by one of the two railway companies, A and B. A string S of length 3 represents which company operates each station. If S_i is A, Company A operates Station i; if S_i is B, Company B operates Station i.

To improve the transportation condition, for each pair of a station operated by Company A and one operated by Company B, there will be a bus service connecting them.

Determine if there is a pair of stations that will be connected by a bus service.

Constraints

Each character of S is A or B.

|S| = 3

Input

Input is given from Standard Input in the following format:

S

Output

If there is a pair of stations that will be connected by a bus service, print Yes; otherwise, print No.

Sample Input 1

ABA

Sample Output 1

Yes

Company A operates Station 1 and 3, while Company B operates Station 2.

There will be a bus service between Station 1 and 2, and between Station 2 and 3, so print Yes.

Sample Input 2

BBA

Sample Output 2

Yes

Company B operates Station 1 and 2, while Company A operates Station 3.

There will be a bus service between Station 1 and 3, and between Station 2 and 3, so print Yes.

Sample Input 3

BBB

Sample Output 3

No

Company B operates all the stations. Thus, there will be no bus service, so print No.

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

#include<iostream>
#include<string>
#include<vector>
#include<numeric>
#include<math.h>
#include<algorithm>
#include<iomanip>
#include<list>
using namespace std;

int main() {

	long A;
	long B;
	cin >> A;
	cin >> B;

	long Amin = 25 * A / 2;
	if (A % 2 != 0) {
		Amin += 1;
	}
	long Amax = 25 * (A + 1) / 2;
	if (A % 2 == 0) {
		Amin -= 1;
	}

	long Bmin = 10 * B;
	long Bmax = 10 * B + 9;

	// 解が存在する場合
	if (Amax >= Bmin&& Bmax >= Amin) {
		long val = max(Amin, Bmin);
		cout << val << endl;
	}
	// 解が存在しない場合
	else {
		cout << -1 << endl;
	}


	return 0;
}