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>
#ifdef ONLINE_JUDGE
#define endl "\n"
#endif
using namespace std;
typedef unsigned long long int lu;
typedef long long int ll;
typedef long double ld;
typedef vector<ll> v;
typedef vector<v> vv;
typedef vector<ld> vld;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef pair<ll, ll> pll;
typedef vector<set<ll>> vsll;
typedef set<pair<ll, ll>> spll;
typedef vector<spll> vspll;
typedef vector<pll> vpll;
const ll MOD = 1e9 + 7;
const v dx = {1, -1, 0, 0};
const v dy = {0, 0, 1, -1};
#define yes cout << "Yes\n";
#define no cout << "No\n";
#define yesno(f) \
    if (f)       \
        yes else no;
#define noo \
    { no return; }
#define round(x, y) ((2 * x + y) / (2 * y))
#define ce(x, y) ((x + y - 1) / y)
#define amax(x, y) \
    if (y > x)     \
        x = y;
#define amin(x, y) \
    if (y < x)     \
        x = y;
#define lcm(x, y) ((x) * (y) / __gcd(x, y))
#define sz(x) (ll) x.size()
#define sq(x) ((x) * (x))

#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

template <typename T = ll>
vector<T> ga(ll n, bool oneIndexed = false) {
    vector<T> a = vector<T>(n + oneIndexed);

    for (ll i = 0; i < n; i++) {
        T p;
        cin >> p;
        a[i + oneIndexed] = p;
    }

    return move(a);
}

template <typename T, typename A>
void pa(vector<T, A> const &a, ll begin = 0, ll end = -1) {
    if (end == -1)
        end = sz(a) - 1;
    for (ll i = begin; i <= end; i++) {
        cout << a[i] << " ";
    }
    cout << endl;
}

void solve() {
    ll a, b, c;
    cin >> a >> b >> c;

    yesno((a == b && b != c) || (c == b && a != c) || (a == c && b != c));
}

int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    // freopen("file", "r", stdin);
    // freopen("file", "w", stdout);
    // ll t;cin >> t;while(t--)
    solve();
    return 0;
}