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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define pa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
#define pb push_back

using ll = long long;
using namespace std;
using pint = pair<int,int>;
using pll = pair<ll,ll>;
const int INFint = 1e9;
const ll INFll = 1e18;
ll MOD=1e9+7;

ll gcd(ll a, ll b) { return __gcd(a,b); } //最大公約数
ll lcm(ll a, ll b) { return a/gcd(a,b)*b; } //最大公倍数

// aのn乗をMODで割りながら計算する
ll modpow(ll a, ll n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % MOD;
    a = a * a % MOD;
    n >>= 1; // right shift
  }
  return res;
}

// MODを法としたaの逆元を計算する
ll modinv(ll a) {
  return modpow(a, MOD - 2);
}

int main(){
  int a,b,c;
  cin>>a>>b>>c;
  if (a == b && b != c){
    cout << "Yes" << endl;
    return 0;
  }
  if (c == b && a != c){
    cout << "Yes" << endl;
    return 0;
  }
  if (a == c && b != c){
    cout << "Yes" << endl;
    return 0;
  }
  cout << "No" << endl;
  return 0;
}