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)

#pragma region header
#include <bitset>
#include <tuple>
#include <cstdint>
#include <cstdio>
#include <cctype>
#include <assert.h>
#include <stdlib.h>
#include <stdio.h>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <limits>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <math.h>
using namespace std;
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rev(i, n) for(int i = (int)(n - 1); i >= 0; i--)
#define rev1(i, n) for(int i = (int)(n); i > 0; i--)
#define pb push_back
#define ts to_string
#define all(v) (v).begin(), (v).end()
#define resort(v) sort((v).rbegin(), (v).rend())
#define vi vector<int>
#define vvi vector<vector<int>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vs vector <string>
#define vvs vector <vector<string>>
using ll = long long;
using P = pair<int, int>;
/* ----------------よく使う数字や配列----------------- */
int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };
constexpr ll mod = 1e9 + 7;
constexpr ll inf = INT32_MAX / 2;
constexpr ll INF = LLONG_MAX / 2;
constexpr long double eps = DBL_EPSILON;
constexpr long double pi = 3.141592653589793238462643383279;
/* ----------------------end----------------------- */

/* --------------------テンプレート------------------ */
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
/* ----------------------end----------------------- */

/* --------------------ライブラリ-------------------- */
ll fact(int i) {       //階乗
    if (i == 0) return 1;
    return (fact(i - 1)) * i % mod;
}
ll gcd(ll a, ll b) {        //最大公約数
    if (b == 0) return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b) {      //最小公倍数
    return a * b / gcd(a, b);
}
int keta(ll n) {        //桁数を求める
    if (n == 0) return 1;
    int count = 0;
    while (n != 0) {
        n /= 10;
        count++;
    }
    return count;
}
bool isprime(ll a)//素数判定、primeならtrue,違うならfalse
{
    if (a < 2) return false;
    else if (a == 2) return true;
    else if (a % 2 == 0) return false;
    double m = sqrt(a);
    for (int i = 3; i <= m; i += 2)
    {
        if (a % i == 0)
        {
            return false;
        }
    }

    // 素数である
    return true;
}
ll ketasum(ll n) {    //各桁の和
    ll sum = 0;
    while (n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}
/* ----------------------end----------------------- */
#pragma endregion
signed main() {
    vi A(3);
    rep(i, 3) {
        cin >> A.at(i);
    }
    int sum = 0;
    for (int x : A) {
        sum += x;
    }
    bool a = false;
    rep(i, 3) {
        rep(j, 3) {
            if (i < j && A.at(i)==A.at(j) &&A.at(i)*3!=sum) {
                a = true;
            }
        }
    }
    if (a) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
    
}