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)

// When you are hungry for something and you're determined and you're focused 
// and you're committed to something it's hard to surrender, isn't it?

#include <bits/stdc++.h>
using namespace std;
#define all(x) (x).begin(), (x).end()
#define endl "\n"
#define ll long long 
#define int ll
#define mod 1000000007
#define mod1 1000000006
#define modi 998244353
#define setbit(x) __builtin_popcount(x)
#define ff first
#define ss second
#define pb push_back
#define fff(i,n) for(int i=0;i<n;i++)
vector<ll> vi[300005];
int color[100005];
ll pa[500005];
ll si[500005];
#define pb push_back
bool visi[1000005]={false};
bool is[100005];
ll fac[10000001];
ll prime[400005];
ll bit[100005];
const long double eps = 1.0e-14;      // allowable error
#define fio ios_base::sync_with_stdio(false); cin.tie(NULL);
 
 
void fact(){
    fac[0]=1;
    for(int i=1;i<=10000000;i++){
        fac[i]=(fac[i-1]*i)%mod;
    }
    return ;
}
void sieve(){
    prime[0]=0;
    prime[1]=0;
    for(int i=2;i<400005;i++){
        prime[i]=1;
    }
    for(int i=2;i*i<400005;i++){
        if(prime[i]==1){
            for(int j=i*i;j<400005;j+=i){
                prime[j]=0;
            }
        }
    }
    
}
 
ll power(ll n , ll p ){
    if(p==0)
        return 1;
    
    ll res = power((n*n)%mod,p/2);
    if(p%2==1)
        res = (n*res)%mod;
    
    return res;
}
 
 
void upd(ll i,ll val,ll n) {
    for(;i<=n;i+=i&(-i))
        bit[i]+=val;
}
 
ll qry(ll i) {
    ll sum=0;
    for(;i>0;i-=i&(-i))
        sum+=bit[i];
    return sum;
}
 
 
ll fin(ll u) {
    if(pa[u]==u)
        return u;
    return fin(pa[u]);
}
 
void uni(ll u,ll v) {
    ll p1=fin(u);
    ll p2=fin(v);
    if(p1==p2)
        return;
    if(si[p1]<si[p2])
        swap(p1,p2);
    si[p1]+=si[p2];
    pa[p2]=p1;
    return;
}
 
int32_t main() {
    fio
    
    int a,b,c;
    cin >> a >> b >> c;
    
    if((a==b and b!=c) or (a==c  and a!=b) or (b==c and a!=b))
    cout  << "Yes" ;
    else
    cout << "No";
        
    return 0;
    
}