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>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vint;
typedef vector<double> vdbl;
typedef vector<ll> vll;
typedef vector<string> vstr;
typedef vector<vector<int> > vvint;
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;
typedef priority_queue<int, vector<int>, greater<int> > spqint; //小さい順に取り出し
typedef priority_queue<ll, vector<ll>, greater<ll> > spqll;     //小さい順に取り出し
typedef priority_queue<int, vector<int>, less<int> > bpqint;    //大きい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll> > bpqll;        //大きい順に取り出し

#define REP(i, n) for (int(i) = 0; (i) < (int)(n); i++)
#define FOR(i, a, b) for (int(i) = a; (i) < (int)b; i++)
#define IREP(i, v) for (auto i = v.begin(); i != v.end(); i++)
#define MOD 1000000007
#define NIL -1
#define FI first
#define SE second
#define MP make_pair
#define PB push_back
#define PF push_front
#define TS to_string
#define BS binary_search
#define LB lower_bound
#define UB upper_bound
#define NP next_permutation
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) (ll) x.size()
#define SP(x) setprecision((ll)x)

const int INF = 1e9;
const ll LINF = 1e18;
const double EPS = 1e-9;
const double PI = M_PI;
const int dx[8] = {1, 0, -1, 0, 1, -1, 1, -1};
const int dy[8] = {0, 1, 0, -1, -1, 1, 1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }     //最小公倍数

void yes()
{
    cout << "Yes" << endl;
}
void no()
{
    cout << "No" << endl;
}
//-----------------------------------------

//-----------------------------------------
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int a, b, c  ;
    cin >> a >> b >> c ;
    if(a==b && b!=c){
        yes() ;
    }
    else if(a!=b && b== c){
        yes() ;
    }
    else{
        no() ;
    }


    

    return 0;
}