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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); ++i)
#define all(x) x.begin(),x.end()
#define ln '\n'
const double PI = acos(-1.0);
const long long MOD = 1000000007LL;
//const long long MOD = 998244353LL;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true;} return false; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true;} return false; }
///////////////////////////////////////////////////////////////////////////////////////////////////

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int A,B,C; cin >> A >> B >> C;
    bool ok = true;
    if (A==B && B==C) ok = true;
    else if (A==B ||B==C||C==A) ok = false;
    cout << (ok ? "No" : "Yes") << ln;
}