Problem:
Score: 100 points

Problem Statement

A triple of numbers is said to be poor when two of those numbers are equal but the other number is different from those two numbers.

You will be given three integers A, B, and C. If this triple is poor, print Yes; otherwise, print No.

Constraints

A, B, and C are all integers between 1 and 9 (inclusive).

Input

Input is given from Standard Input in the following format:

A B C

Output

If the given triple is poor, print Yes; otherwise, print No.

Sample Input 1

5 7 5

Sample Output 1

Yes

A and C are equal, but B is different from those two numbers, so this triple is poor.

Sample Input 2

4 4 4

Sample Output 2

No

A, B, and C are all equal, so this triple is not poor.

Sample Input 3

4 9 6

Sample Output 3

No

Sample Input 4

3 3 4

Sample Output 4

Yes

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

#include "bits/stdc++.h"
using namespace std;
const int MAX = 700000;
const int MOD = 1000000007;

long long  fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x, long long int y){
	return(x + y - 1) / y;

}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;

	if (number2 > number1) {
		m = number2;
		n = number1;
	}
	long long int s = -1;
	while (s != 0) {
		s = m % n;
		m = n;
		n = s;
	}
	return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;
	return m / lcm(m, n) * n;
}
/*逆元計算*/
long long int  modinv(long long a, long long m) {
	long long int b = m, u = 1, v = 0;
	while (b) {
		long long int t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
// index が条件を満たすかどうか
vector<long long int >meguru;
bool isOK(int index, int key) {
	if (meguru[index] >= key) return true;
	else return false;
}
// 汎用的な二分探索のテンプレ
int binary_search(int key) {
	int left = -1; //「index = 0」が条件を満たすこともあるので、初期値は -1
	int right = (int)meguru.size(); // 「index = a.size()-1」が条件を満たさないこともあるので、初期値は a.size()
	/* どんな二分探索でもここの書き方を変えずにできる！ */
	while (right - left > 1) {
		int mid = left + (right - left) / 2;

		if (isOK(mid, key)) right = mid;
		else left = mid;
	}
	/* left は条件を満たさない最大の値、right は条件を満たす最小の値になっている */
	return right;
}
int main() {
	int a, b, c;
	cin >> a >> b >> c;
	if (a == b && b == c) {
		cout << "No";
		return 0;
	}
	if (a != b && b != c && c != a) {
		cout << "No";
		return 0;
	}
	cout << "Yes";
}