stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include<iostream>

using namespace std;



int main(){



 int n,m;

    cin>>n>>m;

    if(n==m){

    cout<<"Yes";}

    

    else if(n!=m){

    cout<<"No";}









}

// Optimized solution to the same problem (faster):
#include<stdio.h>

int main(){

	int N;

	int M;

	scanf("%d",&N);

	scanf("%d",&M);

	

	if(N==M){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	

	

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int main ()

{

    int n,m;

    cin>>n>>m;

    if (n==m)

        cout<<"Yes"<<endl;

    else

        cout<<"No"<<endl;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main(){

	int N;

	int M;

	scanf("%d",&N);

	scanf("%d",&M);

	

	if(N==M){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	

	

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,m;

    cin>>n>>m;

    if(n==m)

        cout<<"Yes"<<endl;

    else cout<<"No"<<endl;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main(){

	int N;

	int M;

	scanf("%d",&N);

	scanf("%d",&M);

	

	if(N==M){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	

	

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int main()

{

    long long n,m;

    cin>>n>>m;

    if(n==m)

    {

        cout<<"Yes"<<endl;

    }

    else

    {

        cout<<"No"<<endl;

    }

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>

int main(){

	int N;

	int M;

	scanf("%d",&N);

	scanf("%d",&M);

	

	if(N==M){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	

	

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//Code by ASIM FOIZE AIMON

//CSE'19;CUET

//avoid copy paste//Hit the frined button

#include <bits/stdc++.h>

using namespace std;

#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);

#define precise cout.precision(10); cout << fixed;

#define endl "\n"

#define int int64_t

#define ll long long

#define yes printf("Yes\n")

#define no printf("No\n")

#define push_back pb

int32_t main()

{

    /*#ifndef ONLINE_JUDGE

        freopen("/media/AIMON/SSDVol/Progs/CodeForces/input.txt", "r", stdin);

        freopen("/media/AIMON/SSDVol/Progs/CodeForces/output.txt", "w", stdout);

    #endif*/

    int i,t,n,m;

    cin>>n>>m;

    if(n==m)

        yes;

    else

        no;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main(){

	int N;

	int M;

	scanf("%d",&N);

	scanf("%d",&M);

	

	if(N==M){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	

	

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

#define gcd(a,b)   __gcd(a,b)

#define lcm(a,b)   (a*b)/gcd

#define rep(i,n)   for(int i=0;i<n;i++)

#define Rep(i,n)   for(int i=1;i<=n;i++)

#define REP(i,j,n) for(int i=j;i<=n;i++)

#define fast   ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)





int main()

{ 

	fast;

	ll n,m;

	cin>>n>>m;

	if(n==m){

		cout<<"Yes"<<endl;

	}

	else cout<<"No"<<endl;

	return 0;



}

// Optimized solution to the same problem (faster):
#include<stdio.h>

int main(){

	int N;

	int M;

	scanf("%d",&N);

	scanf("%d",&M);

	

	if(N==M){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	

	

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;



int main()

{

    int a, b;

    cin >> a >> b;

    if(a == b)

        cout << "Yes" << endl;

    else if(a != b)

        cout << "No" << endl;





    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>



using namespace std;



int main()

{

    int n,m;

    cin >> n >> m;

    if(n == m)

        cout <<"Yes" << endl;

    else

        cout <<"No" << endl;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;

ll n,q,c,up,low,t,x,y,arr[4],a,b;

string str,str1;

bool f;



int main()

{

    cin >> a >> b;

    if (a==b)

        cout << "Yes" << endl;

    else

        cout << "No\n";

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int main()

{

    int M,N;

    cin>>N>>M;

    if(M==N)

    {

        cout<<"Yes"<<endl;

    }

    else

    {

        cout<<"No"<<endl;

    }

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



int main()

{

    int n, m;

    cin>> n >> m;



    if(n == m)

        cout<< "Yes\n";



    else

        cout<< "No\n";



    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
//BISMILLAHIR RAHMANIR RAHIM

//RHYTHM_086_NS_01

//ALAN_IS_ON....







#include <bits/stdc++.h>



//define



#define                boost_                         ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define                iter_(i,n)                     for (int i = 0; i < int(n); i++)

#define                for_n(i, n)                    for (int i = 1; i <= int(n); i++)

#define                pb                             push_back

#define                mp                             make_pair

#define                itr                            ::iterator

#define                s_sort(s)                      sort(s.begin(),s.end())

#define                n_sort(a)                      sort(a,a+n)

#define                thank_you                      return 0

#define                thanks                         return

#define                pii                            pair<int,int>

#define                precise_impact                 cout<<setprecision(10)<<fixed;



using namespace std;



//data type

typedef long long int ll;

typedef unsigned long long int ulli;

typedef double db;





//constants

#define PI acos(-1.00)

#define MOD 1000000007



ll bigmod(ll a,ll b){

    if(b==0){

        return 1;

    }

    ll tm=bigmod(a,b/2);

    tm=(tm*tm)%MOD;

    if(b%2==1)

        tm=(tm*a)%MOD;

    return tm;



}

int egcd(int a,int b,int &x,int &y){

    if(a==0){

        x=0;

        y=1;

        return b;

    }

    int x1,y1;

    int d=egcd(b%a,a,x1,y1);

    x=y1-(b/a)*x1;

    y=x1;

    return d;

}



void solve();





int main()

{

    boost_;

    precise_impact;

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    //Here we go...

    int t=1;

    //cin>>t;

    while(t--)

        solve();



    thank_you;

}





void solve()

{

    int n,m;

    cin>>n>>m;

    if(n==m)

        cout<<"Yes"<<endl;

    else

        cout<<"No"<<endl;





}




// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

main()

{

    int n,m;

    cin>>n>>m;

    if(n==m)

        cout<<"Yes";

    else

        cout<<"No";

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,m;

    cin>>n>>m;

     if(n==m){

        cout<<"Yes"<<endl;

     }

     else{

        cout<<"No"<<endl;

     }

  return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

   int N,M;

   cin>>N>>M;

   if(N==M)

   {

       cout<<"Yes"<<endl;

   }

   else

   {

       cout<<"No"<<endl;

   }

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
//...Bismillahir Rahmanir Rahim...

// Code by    Asad Bin Saber



#include <bits/stdc++.h>



using namespace std;



// typedefs...

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<int, pii> piii;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<ll, ll> pll;



// constants...

const double PI = acos(-1);

const int MOD = 1e9+7; // 998244353;

const int MXS = 2e5+5;

const int MXI = 1e9+5;

const ll MXL = 1e15+5;

const int INF = 1e9+5;



// defines...

#define MP make_pair

#define PB push_back

#define fi first

#define se second

#define si(a) scanf("%d", &a)

#define sii(a, b) scanf("%d%d", &a, &b)

#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)

#define sl(a) scanf("%lld", &a)

#define sll(a, b) scanf("%lld%lld", &a, &b)

#define slll(a, b, c) scanf("%lld%lld%lld", &a, &b, &c)

#define sz(x) (int)x.size()

#define all(x) begin(x), end(x)



// chess moves...

// int dx[] = {+0, +0, -1, +1}; ///Up-down, Left-Right

// int dy[] = {+1, -1, +0, +0};

// int dx[] = {+0,+0,+1,-1,-1,+1,-1,+1}; ///King's Move

// int dy[] = {-1,+1,+0,+0,+1,+1,-1,-1};

// int dx[] = {-2, -2, -1, -1,  1,  1,  2,  2}; ///Knight's Move

// int dy[] = {-1,  1, -2,  2, -2,  2, -1,  1};



// structs...

struct hash_pair { 

    template <class T1, class T2> 

    size_t operator()(const pair<T1, T2>& p) const

    { 

        auto hash1 = hash<T1>{}(p.first); 

        auto hash2 = hash<T2>{}(p.second); 

        return hash1 ^ hash2; 

    } 

}; 



// functions...

ll gcd(ll a, ll b) {

    while (b) {

        a %= b;

        swap(a, b);

    }

    return a;

}

ll lcm(ll a, ll b)

{

    return (a/gcd(a, b)*b);

}

ll comb(ll a, ll b)

{

    ll x = max(a-b, b), ans=1;



    for(ll K=a, L=1; K>=x+1; K--, L++){

        ans *= K;

        ans /= L;

    }



    return ans;

}





void task()

{

	// Code here...

	

	int n, m;

	

	cin >> n >> m;

	

	if(n==m) cout << "Yes\n";

	else cout << "No\n";

}



 

int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);

	

    //freopen("input.txt", "r", stdin);

    //freopen("output.txt", "w", stdout);

	

	task();



    return 0;

}




// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);

#define precise cout.precision(10); cout << fixed;

#define endl "\n"

#define int int64_t

#define ll long long





void solve() {

	int n,m;

	cin>>n>>m;

	cout<<((n==m)?"Yes":"No");

	cout<<endl;

}



int32_t main()

{

	fastio;

	precise;





	//int test;cin >> test;while (test--)

	 {solve();}

	return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

 	int N, M;

 	scanf("%d %d", &N, &M);

 	if(N == M){

 		puts("Yes");

	 }

	 else {

	 	puts("No");

	 }

	 return 0;

 }

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;

 

int main() {

	int a,b;

	cin>>a>>b;

	if(a == b) cout<<"Yes\n";

	else cout<<"No\n";

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

       int n,m;

     cin>>n>>m;



       if(n==m)

       cout<<"Yes";



      else

      cout<<"No";

}

// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

	ios_base :: sync_with_stdio(0);

	int n,m ;

	cin >> n >> m;

	if(n==m){

		cout << "Yes" << endl;

	}

	else{

		cout << "No" << endl;

	}

	return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//BISMILLAHIR RAHMANIR RAHIM

//SOTO MEAW

//SOTO SONDESH

#include<bits/stdc++.h>



#define ls ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define ll long long

#define __ <<" "<<

#define case(z) "Case " << z++ << ": "

#define yes cout << "YES" << endl

#define no cout << "NO" << endl

#define quit return 0



using namespace std;



void showmyDS(set<int> ds)

{

    set<int> ::iterator it;

    for (it=ds.begin();it!=ds.end();it++)

    {

        cout << *it << endl ;

    }

    cout << endl;

}

int main()

{

    ls

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    //int t,l=1;cin>>t;while(t>0){

        int n,m;

        cin >> n >> m ;

        if(n==m)

            cout << "Yes" ;

        else

            cout << "No" ;

    //t--;}



    quit;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,m;

    cin>>n>>m;

    if(n==m) cout<<"Yes";

    else cout<<"No";

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):


#include<bits/stdc++.h>

using namespace std;





//defines...

#define ll long long

#define tata return;

#define pb push_back

#define mp make_pair

#define in insert

#define run ios_base::sync_with_stdio(false);cin.tie(NULL);

#define yes cout<<"Yes"<<endl;

#define no cout<<"No"<<endl;

#define vec vector<ll>

#define setl set<ll>

#define mapl map<ll,ll>

#define shunno cout<<0<<endl;

#define pi 2*acos(0.0)1

#define hobena cout<<-1<<endl;









void oka()

{

   ll n,m;

   cin>>n>>m;

   if(n==m)

    yes

    else

    no





}













int main()

{

   //string g="abcd";

    //g.erase(g.begin()+3);

    //cout<<g<<endl;

    //freopen("input.txt", "r", stdin);

    //freopen("output.txt", "w", stdout);



    run

    ll i,t;

    t=1;

   //cin>>t;





    for(i=1;i<=t;i++)

    {





    oka();

    }













}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
///Bismillahir Rahmanir Rahim

#include<bits/stdc++.h>

using namespace std;

#define                          ll                          long long int

ll gcd(int a,int b)

{

    if(b==0)

        return a;

    return gcd(b,a%b);

}



#define                          in(x)                       scanf("%lld",&x)

#define                          in2(x,y)                    scanf("%lld %lld",&x,&y)

#define                          in3(x,y,z)                  scanf("%lld %lld %lld",&x,&y,&z)

#define                          out(x)                      printf("%lld\n",x)

#define                          cy                          printf("YES\n")

#define                          cn                          printf("NO\n")

#define                          inf                         1e18

#define                          neg                         -1e18

#define                          mx                          2111111

#define                          fast                        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define                          fi                          frist

#define                          si                          second

#define                          ce                          cout<<endl

#define                          pb                          push_back

#define                          vc(x)                       x.begin(),x.end()

#define                          pb                          push_back

#define                          pi                          pair<ll,ll>

#define                          debug                       printf("ok\n");

#define                          pg                          pair<ll,ll>,vector<pair<ll,ll> >,greater<pair<ll,ll> >







int main(){

    fast

  ll test  ;

  ll n,m;

  in2(n,m);

  if(n==m)cout<<"Yes"<<endl;

  else cout<<"No"<<endl;

}




// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

	int n,m;

	cin>>n>>m;

   if(n==m) cout<<"Yes"<<endl;

   else cout<<"No"<<endl;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):


#include<bits/stdc++.h>

int main()

{

    int N,M;

    scanf("%d %d",&N,&M);

    if(N==M)

    {

        printf("Yes");

    }

    else

    {

        printf("No");

    }

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main (){

	

	int m, n;

	scanf ("%d %d", &m, &n);

	if (m == n){

		printf ("Yes\n");

	} 

	else {

		printf ("No\n");

	}	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

int main(){

    int n , m;

    std :: cin >> n >> m;

    if (n==m) std :: cout << "Yes";

    else std:: cout << "No";

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

  int n, m;

  scanf("%d %d", &n, &m);

  if(n == m) puts("Yes");

  else puts("No");

  return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



#define fi first

#define se second

#define mp make_pair

#define pb push_back

#define mt make_tuple

#define eb emplace_back

#define fastio() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)



#define LB(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())

#define UB(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())

typedef long long ll;

typedef pair<ll,ll> ii;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef unsigned long long ull;

typedef long double ld; 

typedef tuple<int,int,int> State;



ll mod = 1e9 + 7;



#define citc(tc) int tc; cin>>tc; while(tc--){solve();}

#define each(a,x) for( auto &a:(x))

#define rep(i, n) for (ll i = 0; i < (n); i++)

#define repr(i, n) for (ll i = (n) - 1; i >= 0; i--)

#define rep2(i, l, r) for (ll i = (l); i < (r); i++)

#define rep2r(i, l, r) for (ll i = (r) - 1; i >= (l); i--)

#define range(a) a.begin(), a.end()

#define deb(x) cout<<#x<<" "<<x<<endl;

#define cini(i) int i; cin>>i;

#define cinll(l) ll l; cin>>l;

#define cinai(a, n) vi a(n); rep(i, n) { cin>>a[i]; }

#define cinal(a, n) vl a(n); rep(i, n) { cin>>a[i]; }

#define coutai(a, n) rep(i, n) { cout<<a[i]<<" "; } ent;

#define coutal(a, n) rep(i, n) { cout<<a[i]<<" "; } ent;

#define ent cout<<endl;



void solve()

{

	ll a, b;

	cin>>a>>b;

	if(a == b)

		cout<<"Yes\n";

	else

		cout<<"No\n";

}

int main()

{

	fastio();

	// citc(tc);

	solve();

}

// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

  int n, m;

  scanf("%d %d", &n, &m);

  if(n == m) puts("Yes");

  else puts("No");

  return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;



int main() {

	int n,m;

	cin>>n>>m;

	if(m==n)cout<<"Yes";

	else cout<<"No";

	return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

  int n, m;

  scanf("%d %d", &n, &m);

  if(n == m) puts("Yes");

  else puts("No");

  return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int m,n;

    cin>>n>>m;

    if(n==m)

        cout<<"Yes"<<endl;

    else

        cout<<"No"<<endl;

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

  int n, m;

  scanf("%d %d", &n, &m);

  if(n == m) puts("Yes");

  else puts("No");

  return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):


#include<bits/stdc++.h>

using namespace std;





#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);  ///for fasting input and output using cin,cout(only)

#define endl '\n'



int main()



{

    optimize();

    int n,m;

    cin>>n>>m;

    if( n == m )

        cout<<"Yes"<<endl;

    else cout<<"No"<<endl;

 return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

  int n, m;

  scanf("%d %d", &n, &m);

  if(n == m) puts("Yes");

  else puts("No");

  return 0;

}

// Fastest known solution for this problem: