Problem:
Score : 100 points

Problem Statement

Takahashi is participating in a programming contest, AXC001. He has just submitted his code to Problem A.

The problem has N test cases, all of which must be passed to get an AC verdict.

Takahashi's submission has passed M cases out of the N test cases.

Determine whether Takahashi's submission gets an AC.

Constraints

1 \leq N \leq 100

0 \leq M \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

If Takahashi's submission gets an AC, print Yes; otherwise, print No.

Sample Input 1

3 3

Sample Output 1

Yes

All three test cases have been passed, so his submission gets an AC.

Sample Input 2

3 2

Sample Output 2

No

Only two out of the three test cases have been passed, so his submission does not get an AC.

Sample Input 3

1 1

Sample Output 3

Yes

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repe(i, n) for (int i = 0; i <= (int)(n); i++)
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define ANS cout << ans << endl
#define all(a) a.begin(),a.end()
#define SORT(a) sort(all(a))
#define REVERSE(a) reverse(all(a))
#define KOZYUN(a) sort(all(a),greater<int>())
#define int ll
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORe(i, m, n) for(int i = m;i <= n;i++)
#define VIin(a) for(int i=0; i<a.size(); i++) cin >> a[i]
#define OUT(a) cout << a << endl
#define IN(a) cin >> a
#define INT(a) int a; cin >> a
typedef vector<int> VI;
typedef vector<VI> VVI;
const int MOD=1e9+7;
const long long INF = 1LL << 60;
/*最大公約数*/int gcd(int x, int y) { return y ? gcd(y,x%y) : x;}
/*最小公倍数*/int lcm(int a, int b) { int g = gcd(a, b); return a / g * b; }
/*素数判定*/int isPrime(int n){if(n<2){return 0;}for(int i=2; i*i<=n; i++){if(n%i==0){return 0;}}return 1;}
/*約数列挙*/vector<int> getDivisor(int n){vector<int> v; for(int i=1; i*i<=n; i++){if(n%i==0){v.push_back(i);if(i!=n/i){v.push_back(n/i);}}}sort(v.begin(),v.end()); return v;}
//----------------------------------------------
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n,m; cin >> n >> m;
    if(n == m) Yes;
    else No;

    return 0;
}