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;
#define REP(i,n) for(long long i=0;i<(n);i++)
#define REP1(i,n) for(long long i=1;i<=(n);i++)
#define REP2D(i,j,h,w) for(long long i=0;i<(h);i++) for(long long j=0;j<(w);j++)
#define REP2D1(i,j,h,w) for(long long i=1;i<=(h);i++) for(long long j=1;j<=(w);j++)
#define PER(i,n) for(long long i=((n)-1);i>=0;i--)
#define PER1(i,n) for(long long i=(n);i>0;i--)
#define FOR(i,a,b) for(long long i=(a);i<(b);i++)
#define FORE(i,a,b) for(long long i=(a);i<=(b);i++)
#define ITE(arr) for(auto ite=(arr).begin();ite!=(arr).end();++ite)
#define ALL(a) ((a).begin()),((a).end())
#define RANGE(a) (a),((a)+sizeof(a))
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define YNPRT(b) cout<<((b)?"Yes":"No")<<endl
#define ENTER printf("\n")
#define REV(arr) reverse(ALL(arr))
#define PRT(a) cout<<(a)<<endl

#ifdef DEBUG
#define DBPRT(a) cout << "[Debug] - " << #a << " : " << (a) << endl
#define DBSTART if(1){
#define DBEND }
#define PRTLST(arr,num) REP(_i,num) cout<<_i<<" - "<<arr[_i]<<endl;
#define PRTLST2(arr2,d1,d2) REP(_i,d1) REP(_j,d2) cout<<_i<<","<<_j<<" : "<<arr2[_i][_j]<<endl;
#define PRTLST2D(arr2,d1,d2) do{cout<<"L\t";REP(_i,d2) cout<<_i<<"\t"; cout<<endl; REP(_i,d1){cout<<_i<<"\t";REP(_j,d2){cout<<arr2[_i][_j]<<"\t";}cout<<endl;}}while(0);
#else
#define DBPRT(a) if(0){}
#define DBSTART if(0){
#define DBEND }
#define PRTLST(arr,num) if(0){}
#define PRTLST2(arr2,d1,d2) if(0){}
#define PRTLST2D(arr2,d1,d2) if(0){}
#endif

#define TOSUM(arr,sum,n) {sum[0]=arr[0];REP1(i,n-1) sum[i]=sum[i-1]+arr[i];}

#define MIN(target,v1) (target)=min(target,v1)
#define MAX(target,v1) (target)=max(target,v1) 
#define P1 first
#define P2 second
#define PB push_back
#define UB upper_bound
#define LB lower_bound
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const int INF_INT = 2147483647;
const ll INF_LL = 9223372036854775807LL;
const ull INF_ULL = 18446744073709551615Ull;
const ll P = 92540646808111039LL;
const int Move[4][2] = {-1,0,0,1,1,0,0,-1};
const int Move_[8][2] = {-1,0,-1,-1,0,1,1,1,1,0,1,-1,0,-1,-1,-1};
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const pair<T1,T2>& p) {os<<"( "<<p.P1<<" , "<<p.P2<<" )";return os;}
template<typename T>
ostream& operator<<(ostream& os,const vector<T>& v) {ITE(v) os << (ite-v.begin()) << " : " << *ite <<endl;return os;}
template<typename T>
ostream& operator<<(ostream& os,const set<T>& v) {os<<" { ";ITE(v) {os<<*ite;if(ite!=--v.end()){os<<" , ";}} os<<" } ";return os;}
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const map<T1,T2>& m) {ITE(m) {os<<ite->P1<<"\t\t|->\t\t"<<ite->P2<<endl;} return os;}

//---------------------
#define MAXN 100000
//---------------------

ll n,m;


int main(){
	cin >> n >> m;
	YNPRT(n==m);
	return 0;
}