stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<bits/stdc++.h>

#define ll long long

#define ull unsinged long long

#define ld long double

#define mem(a,b) memset(a,b,sizeof(a))

#define inf (1LL<<62)

#define sf scanf

#define pf printf

#define repi(i,a,b) for(int i=a;i<=b;++i)

#define rep(i,a,b) for(ll i=a;i<=b;++i)

#define depi(i,a,b) for(int i=a;i>=b;--i)

#define dep(i,a,b) for(ll i=a;i>=b;--i)

#define mod 1000000007



using namespace std;



int ans[100005];



int main()

{

    //std::ios::sync_with_stdio(false);

    int n,k,q,s;

    mem(ans,0);

    sf("%d%d%d",&n,&k,&q);

    int m=q-k;

    for(int i=1;i<=q;++i)

    {

        sf("%d",&s);

        ++ans[s];

    }

    for(int i=1;i<=n;++i)

    {

        if(ans[i]<=m)

            pf("No\n");

        else

            pf("Yes\n");

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.373886237334598x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,m,i,j,sum = 0,q;

    cin>>n>>m>>q;

    map<int,int>mp;

    vector<int>v;

    int a[q];

    for(i = 0;i < q;i++)

    {

        cin>>a[i];

       mp[a[i]]++;

    }

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

    {

        j = mp[i] + m - q;

        if(j>0)

            cout<<"Yes\n";

        else

            cout<<"No\n";

    }



    return 0;

}
// Slower C++ solution:
#include<iostream>

#include<cstdio>

#include<algorithm>

#include <cstring>

using namespace std;

int main()

{

    int t=1,i=0,n,q,a[100005],N[100005]= {0},maxl=0,d;

    long long k;

    cin>>n>>k>>q;

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

    {

        cin>>a[i];

        N[a[i]]++;

    }

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

        if(N[i]>N[maxl])

            maxl=i;

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

        if(a[i]!=maxl)

            --k;

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

        if(N[i]+k-N[maxl]<=0)

            cout<<"No\n";

        else

            cout<<"Yes\n";

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.393645019199654x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,m,i,j,sum = 0,q;

    cin>>n>>m>>q;

    map<int,int>mp;

    vector<int>v;

    int a[q];

    for(i = 0;i < q;i++)

    {

        cin>>a[i];

       mp[a[i]]++;

    }

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

    {

        j = mp[i] + m - q;

        if(j>0)

            cout<<"Yes\n";

        else

            cout<<"No\n";

    }



    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 10;



ll s[MAXN];



int main() {

	ios::sync_with_stdio(false);

	ll n, k, q, a;

	while(cin >> n >> k >> q) {

		memset(s, 0, sizeof(s));

		for(int i = 1; i <= q; i++) {

			cin >> a;

			s[a]++;

		}

		for(int i = 1; i <= n; i++) {

			if(s[i] > q - k) {

				cout << "Yes\n";

			} else {

				cout << "No\n";

			}

		}

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8343827192817188x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,m,i,j,sum = 0,q;

    cin>>n>>m>>q;

    map<int,int>mp;

    vector<int>v;

    int a[q];

    for(i = 0;i < q;i++)

    {

        cin>>a[i];

       mp[a[i]]++;

    }

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

    {

        j = mp[i] + m - q;

        if(j>0)

            cout<<"Yes\n";

        else

            cout<<"No\n";

    }



    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define ll long long

#define ull unsinged long long

#define ld long double

#define mem(a,b) memset(a,b,sizeof(a))

#define inf (1LL<<62)

#define sf scanf

#define pf printf

#define repi(i,a,b) for(int i=a;i<=b;++i)

#define rep(i,a,b) for(ll i=a;i<=b;++i)

#define depi(i,a,b) for(int i=a;i>=b;--i)

#define dep(i,a,b) for(ll i=a;i>=b;--i)

#define mod 1000000007



using namespace std;



int ans[100005];



int main()

{

    //std::ios::sync_with_stdio(false);

    int n,k,q,s;

    mem(ans,0);

    sf("%d%d%d",&n,&k,&q);

    int m=q-k;

    for(int i=1;i<=q;++i)

    {

        sf("%d",&s);

        ++ans[s];

    }

    for(int i=1;i<=n;++i)

    {

        if(ans[i]<=m)

            pf("No\n");

        else

            pf("Yes\n");

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3895713671994414x):


#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int N, K, Q; cin >> N >> K >> Q; int A, P[N];

    for (int i = 1; i <= N; i++) P[i]  = K;

    for (int i = 0; i <  Q; i++) {cin >> A; ++P[A];}

    for (int i = 1; i <= N; i++) P[i] -= Q;

    for (int i = 1; i <= N; i++) {

        if (P[i] > 0) cout << "Yes" << endl;

        else          cout << "No"  << endl;

    }



    return 0;

}
// Slower C++ solution:
#include<iostream>

#include<cstdio>

#include<algorithm>

#include <cstring>

using namespace std;

int main()

{

    int t=1,i=0,n,q,a[100005],N[100005]= {0},maxl=0,d;

    long long k;

    cin>>n>>k>>q;

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

    {

        cin>>a[i];

        N[a[i]]++;

    }

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

        if(N[i]>N[maxl])

            maxl=i;

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

        if(a[i]!=maxl)

            --k;

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

        if(N[i]+k-N[maxl]<=0)

            cout<<"No\n";

        else

            cout<<"Yes\n";

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4095557274647337x):


#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int N, K, Q; cin >> N >> K >> Q; int A, P[N];

    for (int i = 1; i <= N; i++) P[i]  = K;

    for (int i = 0; i <  Q; i++) {cin >> A; ++P[A];}

    for (int i = 1; i <= N; i++) P[i] -= Q;

    for (int i = 1; i <= N; i++) {

        if (P[i] > 0) cout << "Yes" << endl;

        else          cout << "No"  << endl;

    }



    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 10;



ll s[MAXN];



int main() {

	ios::sync_with_stdio(false);

	ll n, k, q, a;

	while(cin >> n >> k >> q) {

		memset(s, 0, sizeof(s));

		for(int i = 1; i <= q; i++) {

			cin >> a;

			s[a]++;

		}

		for(int i = 1; i <= n; i++) {

			if(s[i] > q - k) {

				cout << "Yes\n";

			} else {

				cout << "No\n";

			}

		}

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8553251600689407x):


#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int N, K, Q; cin >> N >> K >> Q; int A, P[N];

    for (int i = 1; i <= N; i++) P[i]  = K;

    for (int i = 0; i <  Q; i++) {cin >> A; ++P[A];}

    for (int i = 1; i <= N; i++) P[i] -= Q;

    for (int i = 1; i <= N; i++) {

        if (P[i] > 0) cout << "Yes" << endl;

        else          cout << "No"  << endl;

    }



    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define ll long long

#define ull unsinged long long

#define ld long double

#define mem(a,b) memset(a,b,sizeof(a))

#define inf (1LL<<62)

#define sf scanf

#define pf printf

#define repi(i,a,b) for(int i=a;i<=b;++i)

#define rep(i,a,b) for(ll i=a;i<=b;++i)

#define depi(i,a,b) for(int i=a;i>=b;--i)

#define dep(i,a,b) for(ll i=a;i>=b;--i)

#define mod 1000000007



using namespace std;



int ans[100005];



int main()

{

    //std::ios::sync_with_stdio(false);

    int n,k,q,s;

    mem(ans,0);

    sf("%d%d%d",&n,&k,&q);

    int m=q-k;

    for(int i=1;i<=q;++i)

    {

        sf("%d",&s);

        ++ans[s];

    }

    for(int i=1;i<=n;++i)

    {

        if(ans[i]<=m)

            pf("No\n");

        else

            pf("Yes\n");

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3894406727109267x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,k,q,i,j,ind;

    cin>>n>>k>>q;

    int a[q],b[n];

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

        b[i]=k;

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

    {

        cin>>a[i];

        ind=a[i];

        b[ind]+=1;

    }

    for(j=1; j<=n; j++)

    {

        b[j]-=q;

    }



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

    {

        if(b[i]>0)

            cout<<"Yes"<<endl;

        else

            cout<<"No"<<endl;

    }



    return 0;

}
// Slower C++ solution:
#include<iostream>

#include<cstdio>

#include<algorithm>

#include <cstring>

using namespace std;

int main()

{

    int t=1,i=0,n,q,a[100005],N[100005]= {0},maxl=0,d;

    long long k;

    cin>>n>>k>>q;

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

    {

        cin>>a[i];

        N[a[i]]++;

    }

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

        if(N[i]>N[maxl])

            maxl=i;

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

        if(a[i]!=maxl)

            --k;

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

        if(N[i]+k-N[maxl]<=0)

            cout<<"No\n";

        else

            cout<<"Yes\n";

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4094231533708925x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,k,q,i,j,ind;

    cin>>n>>k>>q;

    int a[q],b[n];

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

        b[i]=k;

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

    {

        cin>>a[i];

        ind=a[i];

        b[ind]+=1;

    }

    for(j=1; j<=n; j++)

    {

        b[j]-=q;

    }



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

    {

        if(b[i]>0)

            cout<<"Yes"<<endl;

        else

            cout<<"No"<<endl;

    }



    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 10;



ll s[MAXN];



int main() {

	ios::sync_with_stdio(false);

	ll n, k, q, a;

	while(cin >> n >> k >> q) {

		memset(s, 0, sizeof(s));

		for(int i = 1; i <= q; i++) {

			cin >> a;

			s[a]++;

		}

		for(int i = 1; i <= n; i++) {

			if(s[i] > q - k) {

				cout << "Yes\n";

			} else {

				cout << "No\n";

			}

		}

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8551506596592837x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,k,q,i,j,ind;

    cin>>n>>k>>q;

    int a[q],b[n];

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

        b[i]=k;

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

    {

        cin>>a[i];

        ind=a[i];

        b[ind]+=1;

    }

    for(j=1; j<=n; j++)

    {

        b[j]-=q;

    }



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

    {

        if(b[i]>0)

            cout<<"Yes"<<endl;

        else

            cout<<"No"<<endl;

    }



    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define ll long long

#define ull unsinged long long

#define ld long double

#define mem(a,b) memset(a,b,sizeof(a))

#define inf (1LL<<62)

#define sf scanf

#define pf printf

#define repi(i,a,b) for(int i=a;i<=b;++i)

#define rep(i,a,b) for(ll i=a;i<=b;++i)

#define depi(i,a,b) for(int i=a;i>=b;--i)

#define dep(i,a,b) for(ll i=a;i>=b;--i)

#define mod 1000000007



using namespace std;



int ans[100005];



int main()

{

    //std::ios::sync_with_stdio(false);

    int n,k,q,s;

    mem(ans,0);

    sf("%d%d%d",&n,&k,&q);

    int m=q-k;

    for(int i=1;i<=q;++i)

    {

        sf("%d",&s);

        ++ans[s];

    }

    for(int i=1;i<=n;++i)

    {

        if(ans[i]<=m)

            pf("No\n");

        else

            pf("Yes\n");

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.389973128338958x):
#include<bits/stdc++.h>

#define ll long long int

using namespace std;

int arr[100008];

int main(){

 	int n,s,m;

 	cin>>n>>s>>m;

	 for(int i=1;i<=m;++i)

	 {

	 	int d;

	 	cin>>d;

	 	arr[d]++;

	  } 

	  for(int i=1;i<=n;++i)

	  {

	  	if(s+arr[i]>m)cout<<"Yes"<<endl;

	  	else cout<<"No"<<endl;

	  }

}
// Slower C++ solution:
#include<iostream>

#include<cstdio>

#include<algorithm>

#include <cstring>

using namespace std;

int main()

{

    int t=1,i=0,n,q,a[100005],N[100005]= {0},maxl=0,d;

    long long k;

    cin>>n>>k>>q;

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

    {

        cin>>a[i];

        N[a[i]]++;

    }

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

        if(N[i]>N[maxl])

            maxl=i;

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

        if(a[i]!=maxl)

            --k;

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

        if(N[i]+k-N[maxl]<=0)

            cout<<"No\n";

        else

            cout<<"Yes\n";

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4099632666013668x):
#include<bits/stdc++.h>

#define ll long long int

using namespace std;

int arr[100008];

int main(){

 	int n,s,m;

 	cin>>n>>s>>m;

	 for(int i=1;i<=m;++i)

	 {

	 	int d;

	 	cin>>d;

	 	arr[d]++;

	  } 

	  for(int i=1;i<=n;++i)

	  {

	  	if(s+arr[i]>m)cout<<"Yes"<<endl;

	  	else cout<<"No"<<endl;

	  }

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 10;



ll s[MAXN];



int main() {

	ios::sync_with_stdio(false);

	ll n, k, q, a;

	while(cin >> n >> k >> q) {

		memset(s, 0, sizeof(s));

		for(int i = 1; i <= q; i++) {

			cin >> a;

			s[a]++;

		}

		for(int i = 1; i <= n; i++) {

			if(s[i] > q - k) {

				cout << "Yes\n";

			} else {

				cout << "No\n";

			}

		}

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8558615827155769x):
#include<bits/stdc++.h>

#define ll long long int

using namespace std;

int arr[100008];

int main(){

 	int n,s,m;

 	cin>>n>>s>>m;

	 for(int i=1;i<=m;++i)

	 {

	 	int d;

	 	cin>>d;

	 	arr[d]++;

	  } 

	  for(int i=1;i<=n;++i)

	  {

	  	if(s+arr[i]>m)cout<<"Yes"<<endl;

	  	else cout<<"No"<<endl;

	  }

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define ll long long

#define ull unsinged long long

#define ld long double

#define mem(a,b) memset(a,b,sizeof(a))

#define inf (1LL<<62)

#define sf scanf

#define pf printf

#define repi(i,a,b) for(int i=a;i<=b;++i)

#define rep(i,a,b) for(ll i=a;i<=b;++i)

#define depi(i,a,b) for(int i=a;i>=b;--i)

#define dep(i,a,b) for(ll i=a;i>=b;--i)

#define mod 1000000007



using namespace std;



int ans[100005];



int main()

{

    //std::ios::sync_with_stdio(false);

    int n,k,q,s;

    mem(ans,0);

    sf("%d%d%d",&n,&k,&q);

    int m=q-k;

    for(int i=1;i<=q;++i)

    {

        sf("%d",&s);

        ++ans[s];

    }

    for(int i=1;i<=n;++i)

    {

        if(ans[i]<=m)

            pf("No\n");

        else

            pf("Yes\n");

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3781141985994345x):
	#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

using namespace __gnu_pbds;



#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'

#define mset(d,val) memset(d,val,sizeof(d))

#define setp(x) cout<<fixed<<setprecision(x)

#define forn(i,a,b) for(int i=(a);i<(b);i++)

#define fore(i,a,b) for(int i=(a);i<=(b);i++)

#define pb push_back

#define F first

#define S second

#define pqueue priority_queue

#define fbo find_by_order

#define ook order_of_key

typedef long long ll;

typedef pair<ll,ll> ii;

typedef vector<ll> vi;

typedef vector<ii> vii;

typedef long double ld;

typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;

void amin(ll &a, ll b){ a=min(a,b); }

void amax(ll &a, ll b){ a=max(a,b); }

void SD(int t=0){ cout<<"PASSED "<<t<<endl; }

const ll INF = ll(1e18);

const int MOD = 998244353;



const int MAXN = 100005;



int main()

{

	ios_base::sync_with_stdio(0); cin.tie(0);

	

	int n,K,Q; cin>>n>>K>>Q;

	

	ll a[n]{};

	forn(i,0,n) a[i]=K-Q;

	forn(i,0,Q){

		int x; cin>>x; x--;

		a[x]++;

	}

	

	forn(i,0,n){

		if(a[i]>0) cout<<"Yes\n";

		else cout<<"No\n";

	}

	

	return 0;

}
// Slower C++ solution:
#include<iostream>

#include<cstdio>

#include<algorithm>

#include <cstring>

using namespace std;

int main()

{

    int t=1,i=0,n,q,a[100005],N[100005]= {0},maxl=0,d;

    long long k;

    cin>>n>>k>>q;

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

    {

        cin>>a[i];

        N[a[i]]++;

    }

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

        if(N[i]>N[maxl])

            maxl=i;

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

        if(a[i]!=maxl)

            --k;

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

        if(N[i]+k-N[maxl]<=0)

            cout<<"No\n";

        else

            cout<<"Yes\n";

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.3979337856185823x):
	#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

using namespace __gnu_pbds;



#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'

#define mset(d,val) memset(d,val,sizeof(d))

#define setp(x) cout<<fixed<<setprecision(x)

#define forn(i,a,b) for(int i=(a);i<(b);i++)

#define fore(i,a,b) for(int i=(a);i<=(b);i++)

#define pb push_back

#define F first

#define S second

#define pqueue priority_queue

#define fbo find_by_order

#define ook order_of_key

typedef long long ll;

typedef pair<ll,ll> ii;

typedef vector<ll> vi;

typedef vector<ii> vii;

typedef long double ld;

typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;

void amin(ll &a, ll b){ a=min(a,b); }

void amax(ll &a, ll b){ a=max(a,b); }

void SD(int t=0){ cout<<"PASSED "<<t<<endl; }

const ll INF = ll(1e18);

const int MOD = 998244353;



const int MAXN = 100005;



int main()

{

	ios_base::sync_with_stdio(0); cin.tie(0);

	

	int n,K,Q; cin>>n>>K>>Q;

	

	ll a[n]{};

	forn(i,0,n) a[i]=K-Q;

	forn(i,0,Q){

		int x; cin>>x; x--;

		a[x]++;

	}

	

	forn(i,0,n){

		if(a[i]>0) cout<<"Yes\n";

		else cout<<"No\n";

	}

	

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 10;



ll s[MAXN];



int main() {

	ios::sync_with_stdio(false);

	ll n, k, q, a;

	while(cin >> n >> k >> q) {

		memset(s, 0, sizeof(s));

		for(int i = 1; i <= q; i++) {

			cin >> a;

			s[a]++;

		}

		for(int i = 1; i <= n; i++) {

			if(s[i] > q - k) {

				cout << "Yes\n";

			} else {

				cout << "No\n";

			}

		}

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8400278002725985x):
	#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

using namespace __gnu_pbds;



#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'

#define mset(d,val) memset(d,val,sizeof(d))

#define setp(x) cout<<fixed<<setprecision(x)

#define forn(i,a,b) for(int i=(a);i<(b);i++)

#define fore(i,a,b) for(int i=(a);i<=(b);i++)

#define pb push_back

#define F first

#define S second

#define pqueue priority_queue

#define fbo find_by_order

#define ook order_of_key

typedef long long ll;

typedef pair<ll,ll> ii;

typedef vector<ll> vi;

typedef vector<ii> vii;

typedef long double ld;

typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;

void amin(ll &a, ll b){ a=min(a,b); }

void amax(ll &a, ll b){ a=max(a,b); }

void SD(int t=0){ cout<<"PASSED "<<t<<endl; }

const ll INF = ll(1e18);

const int MOD = 998244353;



const int MAXN = 100005;



int main()

{

	ios_base::sync_with_stdio(0); cin.tie(0);

	

	int n,K,Q; cin>>n>>K>>Q;

	

	ll a[n]{};

	forn(i,0,n) a[i]=K-Q;

	forn(i,0,Q){

		int x; cin>>x; x--;

		a[x]++;

	}

	

	forn(i,0,n){

		if(a[i]>0) cout<<"Yes\n";

		else cout<<"No\n";

	}

	

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define ll long long

#define ull unsinged long long

#define ld long double

#define mem(a,b) memset(a,b,sizeof(a))

#define inf (1LL<<62)

#define sf scanf

#define pf printf

#define repi(i,a,b) for(int i=a;i<=b;++i)

#define rep(i,a,b) for(ll i=a;i<=b;++i)

#define depi(i,a,b) for(int i=a;i>=b;--i)

#define dep(i,a,b) for(ll i=a;i>=b;--i)

#define mod 1000000007



using namespace std;



int ans[100005];



int main()

{

    //std::ios::sync_with_stdio(false);

    int n,k,q,s;

    mem(ans,0);

    sf("%d%d%d",&n,&k,&q);

    int m=q-k;

    for(int i=1;i<=q;++i)

    {

        sf("%d",&s);

        ++ans[s];

    }

    for(int i=1;i<=n;++i)

    {

        if(ans[i]<=m)

            pf("No\n");

        else

            pf("Yes\n");

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3737586831652546x):




#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

using namespace __gnu_pbds;

 

#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'

#define mset(d,val) memset(d,val,sizeof(d))

#define setp(x) cout<<fixed<<setprecision(x)

#define forn(i,a,b) for(int i=a;i<b;i++)

#define fore(i,a,b) for(int i=a;i<=b;i++)

#define pb push_back

#define mp make_pair

#define F first

#define S second

#define INF ll(1e18)

#define MOD 998244353

#define pqueue priority_queue

#define fbo find_by_order

#define ook order_of_key

typedef long long ll;

typedef pair<ll,ll> ii;

typedef vector<ll> vi;

typedef vector<ii> vii;

typedef long double ld;

typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;

 

#define MAXN 1000000007

 

 

int main() {

	ios_base::sync_with_stdio(0); cin.tie(0);

	ll n,k,q,t;

	cin>>n>>k>>q;

	ll arr[n];

	for (ll i=0;i<n;i++) {

		arr[i]=0;

	}

	for (ll i=0;i<q;i++) {

		cin>>t;

		arr[t-1]+=1;

	}

	for (ll i=0;i<n;i++) {

		if (arr[i]<=q-k) {

			cout<<"No"<<endl;

		}

		else {

			cout<<"Yes"<<endl;

		}

	}

}
// Slower C++ solution:
#include<iostream>

#include<cstdio>

#include<algorithm>

#include <cstring>

using namespace std;

int main()

{

    int t=1,i=0,n,q,a[100005],N[100005]= {0},maxl=0,d;

    long long k;

    cin>>n>>k>>q;

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

    {

        cin>>a[i];

        N[a[i]]++;

    }

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

        if(N[i]>N[maxl])

            maxl=i;

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

        if(a[i]!=maxl)

            --k;

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

        if(N[i]+k-N[maxl]<=0)

            cout<<"No\n";

        else

            cout<<"Yes\n";

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.393515630588026x):




#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

using namespace __gnu_pbds;

 

#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'

#define mset(d,val) memset(d,val,sizeof(d))

#define setp(x) cout<<fixed<<setprecision(x)

#define forn(i,a,b) for(int i=a;i<b;i++)

#define fore(i,a,b) for(int i=a;i<=b;i++)

#define pb push_back

#define mp make_pair

#define F first

#define S second

#define INF ll(1e18)

#define MOD 998244353

#define pqueue priority_queue

#define fbo find_by_order

#define ook order_of_key

typedef long long ll;

typedef pair<ll,ll> ii;

typedef vector<ll> vi;

typedef vector<ii> vii;

typedef long double ld;

typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;

 

#define MAXN 1000000007

 

 

int main() {

	ios_base::sync_with_stdio(0); cin.tie(0);

	ll n,k,q,t;

	cin>>n>>k>>q;

	ll arr[n];

	for (ll i=0;i<n;i++) {

		arr[i]=0;

	}

	for (ll i=0;i<q;i++) {

		cin>>t;

		arr[t-1]+=1;

	}

	for (ll i=0;i<n;i++) {

		if (arr[i]<=q-k) {

			cout<<"No"<<endl;

		}

		else {

			cout<<"Yes"<<endl;

		}

	}

}