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 (Clang 3.8.0) (judge verdict: Wrong Answer)


/*
このコード、と～おれ!
Be accepted!
∧＿∧　
（｡･ω･｡)つ━☆・*。
⊂　　 ノ 　　　・゜+.
　しーＪ　　　°。+ *´¨)
 　　　　　　　　　.· ´¸.·*´¨) ¸.·*¨)
		  　　　　　　　　　　(¸.·´ (¸.·'* ☆
					*/

#include <stdio.h>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <vector>
#include <numeric>
#include <iostream>
#include <random>
#include <map>
#include <unordered_map>
#include <queue>
#include <regex>
#include <functional>
#include <complex>
#include <list>
#include <cassert>
#include <iomanip>
#include <set>

#define rep(i, n) for(int i = 0; i < (n); ++i)
#define rep1(i, n) for(int i = 1; i <= (n); ++i)
#define rep2(i, n) for(int i = 2; i < (n); ++i)
#define repr(i, n) for(int i = n; i >= 0; --i)
#define reprm(i, n) for(int i = n - 1; i >= 0; --i)
#define printynl(a) printf(a ? "yes\n" : "no\n")
#define printyn(a) printf(a ? "Yes\n" : "No\n")
#define printYN(a) printf(a ? "YES\n" : "NO\n")
#define printim(a) printf(a ? "possible\n" : "imposible\n")
#define printdb(a) printf("%.50lf\n", a)//少数出力
#define printdbd(a) printf("%.16lf\n", a)//少数出力(桁少なめ)
#define prints(s) printf("%s\n", s.c_str())//string出力
#define all(x) (x).begin(), (x).end()
#define allsum(a, b, c) ((a + b) * c / 2)//等差数列の和、初項,末項,項数
#define pb push_back
#define priq priority_queue
#define rpriq priq<int, vector<int>, greater<int>>
#define deg_to_rad(deg) (((deg)/360.0)*2.0*PI)
#define rad_to_deg(rad) (((rad)/2.0/PI)*360.0)
#define Please return
#define AC 0
#define addf(T) [](T a, T b){return (a + b);}
#define minf(T) [](T a, T b){return min(a, b);}
#define maxf(T) [](T a, T b){return max(a, b);}


using ll = long long;

constexpr int INF = 1073741823;
constexpr int MINF = -1073741823;
constexpr ll LINF = ll(4661686018427387903);
constexpr ll MOD = 1000000007;
const long double PI = acos(-1.0L);

using namespace std;

void scans(string& str) {
	char c;
	str = "";
	scanf("%c", &c);
	if (c == '\n')scanf("%c", &c);
	while (c != '\n' && c != -1) {
		str += c;
		scanf("%c", &c);
	}
}

void scanc(char& str) {
	char c;
	scanf("%c", &c);
	if (c == -1)return;
	while (c == '\n') {
		scanf("%c", &c);
	}
	str = c;
}

double acot(double x) {
	return PI / 2 - atan(x);
}

ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}

ll lcm(ll number1, ll number2) {
	return number1 / gcd(number1, number2) * number2;
}

ll LSB(ll n) { return (n & (-n)); }

/*-----------------------------------------ここからコード-----------------------------------------*/


int main() {

	int a, b, c;
	scanf("%d%d%d", &a, &b, &c);
	printyn((a xor b) + (b xor c) + (c xor a) == 2);
	
	Please AC;
}