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>
#include <bits/extc++.h>
using namespace __gnu_pbds;
using namespace std;
#define testing 1
#define FOR(ii, ss, ee) for(ll ii = (ss); ii <= (ll)(ee); ++ii)
#define DEC(ii, ss, ee) for(ll ii = (ss); ii >= (ll)(ee); --ii)
#define IAMSPEED ios_base::sync_with_stdio(false); cin.tie(0);
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << " , " << #y << "=" << y << endl
#define db3(a,b,c) cerr<<#a<<"="<<a<<","<<#b<<"="<<b<<","<<#c<<"="<<c<<endl
#define dbv(v) cerr << #v << ":"; for (auto it : v) cerr << it << ' '; cerr << endl
#define dbvp(v) cerr << #v << ":"; for (auto it : v) cerr << "{"  << it.f << ',' << it.s << "} "; cerr << endl
#define dba(a,ss,ee) cerr << #a << ":"; FOR(i,ss,ee) cerr << a[i] << ' '; cerr << endl
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define ll long long 
#define pb push_back
#define inf (int)1e9+500
#define oo (ll)1e18
#define all(x) (x).begin(), (x).end()
#define f first
#define s second
#define reach cerr << "hi\n";
typedef pair <int, int> pi;
typedef pair <ll,ll> pl;
typedef pair <pi, int> pii;
typedef tuple<int,int,int> ti3;
string cts(char x) {string s(1,x); return s;}
const int MOD = 1e9 + 7;
template <typename T> void maxself(T& a, T b) {
	a = max(a, b);
}
template <typename T> void minself(T& a, T b) {
	a = min(a, b);
}
#define MAXN
#if !testing
#define cerr if(0)cout
#endif
int32_t main() {
	IAMSPEED
	// set testing to 0 when submitting to judge
	map<int,int> mp;
	int a,b,c;cin>>a>>b>>c;
	mp[a]++;
	mp[b]++;
	mp[c]++;
	if(mp[a]==2||mp[b]==2||mp[c]==2)cout<<"Yes";
	else cout << "No";
}