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: Wrong Answer)

/* #include<bits/stdc++.h>
#define long long long
using namespace std;
const int N=1e6,inf=1e9;
double aa[N+2],qm[N+2];
int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);
  int n,k;cin>>n>>k;
  for(int i=1;i<=n;i++)
  {
    int x;cin>>x;
    aa[i]=(1.0*((x*(x+1))/2))/(1.0*x);
    qm[i]=qm[i-1]+aa[i];
  }
  double mx=0;
  for(int i=k;i<=n;i++)
    mx=max(mx,qm[i]-qm[i-k]);
  cout<<setprecision(10)<<fixed<<mx<<endl;

  return 0;
} */
/*
#include<bits/stdc++.h>
using namespace std;
#define FasterIO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

typedef unsigned long long ull;
typedef long long ll;
typedef double db;

#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define ff first
#define ss second
#define mk make_pair
#define pi pair<int,int>
#define vi vector<int>
#define pL pair<ll,ll>
#define vL vector<ll>
#define vp vector<pi>
#define vpL vector<pL>

const int MX=300000;

double a[MX+5];

int main()
{
    FasterIO;

    int n, m, k;
    cin>>n>>k;
    for(int i=1; i<=n; i++)
    {
        ll x;
        cin>>x; a[i]=((x*(x+1))/2)/(x*1.0);
    }
    double mx=0, sm=0;
    for(int i=1; i<=k; i++) sm+=a[i];
    mx=sm;
    for(int i=k+1; i<=n; i++)
    {
        sm+=a[i]; sm-=a[i-k]; mx=max(mx, sm);
    }
    cout<<setprecision(10)<<fixed<<mx<<endl;
    return 0;
} */
/*
#include<bits/stdc++.h>
using namespace std;
#define FasterIO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

typedef unsigned long long ull;
typedef long long ll;
typedef double db;

#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define ff first
#define ss second
#define mk make_pair
#define pi pair<int,int>
#define vi vector<int>
#define pL pair<ll,ll>
#define vL vector<ll>
#define vp vector<pi>
#define vpL vector<pL>

const int MX=3000;

ll dp[MX][2][10], n;
string s;
ll call(int p, int f, int k)
{
    if(p==n) return !k;
    ll &ret=dp[p][f][k];
    if(!k)   return ret=1;
    if(~ret) return ret;
    ret=0;
    if(f)
    {
        ret+=call(p+1, 1, k);
        for(int i=1; i<10; i++)
        {
            ret+=call(p+1, 1, k-1);
        }
    }
    else
    {
        for(int i=0; i<s[p]-'0'; i++)
        {
            ret+=call(p+1, 1, (i)?k-1:k);
        }
        ret+=call(p+1, f, (s[p]!='0')?k-1:k);
    }
    return ret;
}

int main()
{
    FasterIO;

    int m, k;
    cin>>s>>k; n=s.size();
    mem(dp, -1);
    ll r1=call(0, 0, k);
    cout<<r1<<endl;
    return 0;
}
*/
/*
#include<bits/stdc++.h>
#define long long long
using namespace std;
const int N=4e6,mod=1e9+7;
int big(int b,int p,int m=mod)
{
  int ret=1;
  while(p){if(p&1)ret=(1LL*ret*b)%m;b=(1LL*b*b)%m;p/=2;}
  return ret;
}
int fac[N+2];
void init(int n)
{
  fac[0]=1;
  for(int i=1;i<=n;i++)fac[i]=(1LL*fac[i-1]*i)%mod;
}
int ncr(int n,int r)
{
  int a=fac[n];a=(1LL*a*big(fac[r],mod-2))%mod;
  return (1LL*a*big(fac[n-r],mod-2))%mod;
}
int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);
  init(N);

  int r1,c1,r2,c2;cin>>r1>>c1>>r2>>c2;

  int ans=0;
  for(int c=c1;c<=c2;c++)
  {
    ans=(ans+ncr(r2+c+1,r2))%mod;
    ans=((ans-ncr(r1+c,r1-1))%mod+mod)%mod;
  }
  cout<<ans<<endl;

  return 0;
} */
#include <bits/stdc++.h>
using ll = long long;
using namespace std;
int main()
{
    int a,b,c; cin >> a >> b>> c;
    if(a == b || b == c || c == a)
    {
        cout << "Yes" << '\n';
        
    }
    else cout << "No" << '\n';
}