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>

#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define size_t unsigned long long
#define ll long long
#define rep(i,a) for(int i=0;i<(a);i++)
#define repr(i,a) for(int i=(int)(a)-1;i>=0;i--)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define FORR(i,a,b) for(int i=(int)(b)-1;i>=a;i--)
#define ALL(a) a.begin(), a.end()
using namespace std;
int si() { int x; scanf("%d", &x); return x; }
long long sl() { long long x; scanf("%lld", &x); return x; }
string ss() { string x; cin >> x; return x; }
void pi(int x) { printf("%d ", x); }
void pl(long long x) { printf("%lld ", x); }
void pd(double x) { printf("%.15f ", x); }
void ps(const string &s) { printf("%s ", s.c_str()); }
void br() { putchar('\n'); }

const ll MOD = 1e9 + 7;
const ll INF = 1e9 + 5;

struct mint {
    int n;
    mint(int n_ = 0) : n(n_) {}
};

mint operator+(mint a, mint b) { a.n += b.n; if (a.n >= MOD) a.n -= MOD; return a; }
mint operator-(mint a, mint b) { a.n -= b.n; if (a.n < 0) a.n += MOD; return a; }
mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; }
mint &operator+=(mint &a, mint b) { return a = a + b; }
mint &operator-=(mint &a, mint b) { return a = a - b; }
mint &operator*=(mint &a, mint b) { return a = a * b; }

typedef pair<int, int> P;

string s,t = "Yes";
ll n,m,a,b;
char c;

int main () {
  cin >> n >> m;
  if(n != m) t = "No";
  cout << t << endl;
}