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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<iostream>
#include<vector>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<string.h>
#include<functional>
#include<limits.h>
#include<stdlib.h>
#include<cmath>
#include<cstring>
#include<set>
#include<climits>
#include<deque>
#include <iomanip>
#include<utility>
using namespace std;

#define intmax INT_MAX
#define lmax LONG_MAX
#define uintmax UINT_MAX
#define ulmax ULONG_MAX
#define llmax LLONG_MAX
#define ll long long
#define rep(i,a,N) for((i)=(a);(i)<(N);(i)++)
#define rrp(i,N,a) for((i)=(N)-1;(i)>=(a);(i)--)
#define llfor ll i,j
#define sc(a) cin>>a
#define pr(a) cout<<a<<endl
#define pY puts("YES")
#define pN puts("NO")
#define py puts("Yes")
#define pn puts("No")
#define pnn printf("\n")
#define all(a) a.begin(),a.end()
#define push(a,b) (a).push_back(b)
#define llvec vector<vector<ll>>
#define charvec vector<vector<char>>
#define size(a,b) (a,vector<ll>(b))
#define llpvec vector<pair<ll,ll>>
#define S 21 
llfor;ll mod=1e9+7;
/*pi*/double pi=acos(-1);
/*繰り上げ除算*/ll cei(ll x,ll y){ll ans=x/y;if(x%y!=0)ans++;return ans;}
/*最大公約数*/ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
/*最小公倍数*/ll lcm(ll x,ll y){return x/gcd(x,y)*y;}
/*n乗*/ll llpow(ll x,ll n){ll ans=1;rep(i,0,n){ans*=x;ans%=mod;}return ans;}
/*階乗*/ll fact(ll x){ll ans=1;rep(i,0,x)ans*=(x-i);return ans;}
/*nCr*/ll ncr(ll n,ll r){ll ans=1,l=2;rep(i,0,min(n-r,r)){ans*=(n-i);ans%=mod;while(ans%l==0&&l<=min(r,n-r)){ans/=l;l++;}}return ans;}
/*nPr*/ll npr(ll n,ll r){return fact(n)/fact(n-r);}
/*primejudge*/bool prime(ll a){if(a<=1)return false;for(i=2;i*i<=a;i++){if(a%i==0)return false;}return true;}//素数の時true
/*Fibonacci数列*/ll fib(ll x){ll fibo[x+10];fibo[0]=1;fibo[1]=1;fibo[2]=1;rep(i,3,x+1)fibo[i]=fibo[i-1]+fibo[i-2];return fibo[x];}
/*桁数*/ll dig(ll n){ll d=1,tmp=n;while(tmp/10>0){tmp/=10;d++;}return d;}
/*行列のコピー*/void matcpy(ll a[S][S],ll b[S][S]){ll t=i;rep(i,0,S)rep(j,0,S)b[i][j]=a[i][j];/*(aをbに)*/i=t;}
/*行列のリセット*/void matrst(ll a[S][S]){ll t=i;rep(i,0,S)rep(j,0,S)a[i][j]=0;i=t;}
/*行列の2乗*/void matpow(ll a[S][S]){ll ans[S][S];matrst(ans);ll t=i,k;rep(i,0,S)rep(j,0,S)rep(k,0,S)ans[i][j]+=a[i][k]*a[k][j];matcpy(ans,a);i=t;}
/*行列の乗算*/void matmul(ll a[S][S],ll b[S][S], ll as[S][S]){ll t=i,k;matrst(as);rep(i,0,S)rep(j,0,S)rep(k,0,S)as[i][j]+=a[i][k]*b[k][j];i=t;}
//ll ans=0;
////////////////////////////////////////////////////////////
int main(){
 ll a,b,c;
 cin>>a>>b>>c;
 if((a-b)*(b-c)*(c-a)==0&&a+b+c!=3*a)py;
 else pn;
return 0;}