stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

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

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

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

		{

			scanf("%lld", &m);

			A[m]++;

		}

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

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 8.96133089407593x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

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

        a[i] = k;

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

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

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

    {

        a[i] -= q;

    }

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

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

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

#include<stdio.h>

#include<algorithm>

#include<stack>

#include<cstring>

#include<cmath>

#include<queue>

#include<set>

using namespace std;

typedef long long ll;

ll a[1111111];

ll b[1111111];

ll c[1111111];



int main()

{

	//ios::sync_with_stdio(false);

	ll n,k,q,x;

	cin>>n>>k>>q;

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

		a[i]=k;

	set<ll> st; 

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

	{

		scanf("%lld",&x);

		a[x-1]++;

	}

	



	for(ll j=0;j<n;j++)

		if(a[j]-q>0)

			cout<<"Yes"<<endl;

		else

			cout<<"No"<<endl;

	return 0;

}




// Faster C++ solution for the same problem (measured speedup: 4.889210132312151x):
#include<stdio.h>



int main()

{

    long long int n,k,q;

    long long int p;

    long long int a[100005];

    scanf("%lld%lld%lld",&n,&k,&q);

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

        a[i] = k;

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

    {

        scanf("%lld",&p);

        a[p] ++ ;

    }

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

    {

        a[i] -= q;

    }

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

    {

        if(a[i] <= 0)

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

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

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

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

    {

        int x;

        cin>>x;

        a[x]++;

    }

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

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4034494644132403x):
#include <bits/stdc++.h>

using namespace std;



int N,K,Q,a;

int arr[100010];

int main(){

	cin>>N>>K>>Q;

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

		arr[i]=K-Q;

	} for(int i=1; i<=Q; i++) {

		cin>>a;

		arr[a]++;

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

		if(arr[i]>0) {

			cout<<"Yes"<<endl;

		} else {

			cout<<"No"<<endl;

		}

	}

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

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

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

{a[i]+=K; a[i]-=Q; }

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

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.1309722686814085x):
#include <bits/stdc++.h>

using namespace std;



int N,K,Q,a;

int arr[100010];

int main(){

	cin>>N>>K>>Q;

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

		arr[i]=K-Q;

	} for(int i=1; i<=Q; i++) {

		cin>>a;

		arr[a]++;

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

		if(arr[i]>0) {

			cout<<"Yes"<<endl;

		} else {

			cout<<"No"<<endl;

		}

	}

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

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

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

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

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

		{

			scanf("%lld", &m);

			A[m]++;

		}

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

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8516034263495442x):
#include <bits/stdc++.h>

using namespace std;



int N,K,Q,a;

int arr[100010];

int main(){

	cin>>N>>K>>Q;

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

		arr[i]=K-Q;

	} for(int i=1; i<=Q; i++) {

		cin>>a;

		arr[a]++;

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

		if(arr[i]>0) {

			cout<<"Yes"<<endl;

		} else {

			cout<<"No"<<endl;

		}

	}

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

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

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

    {

        int x;

        cin>>x;

        a[x]++;

    }

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

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.404001752436814x):
#include<iostream>

using namespace std;

typedef long long ll;

const int maxn=1e5+10;

ll sq[maxn];

int main()

{

	ll n,k,q,m;

	cin>>n>>k>>q;

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

	{

		cin>>m;

		sq[m]++;

	}

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

	{

		if(k-q+sq[i]>0) cout<<"Yes"<<endl;

		else cout<<"No"<<endl;

	}

	return 0;

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

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

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

{a[i]+=K; a[i]-=Q; }

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

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.132991418358783x):
#include<iostream>

using namespace std;

typedef long long ll;

const int maxn=1e5+10;

ll sq[maxn];

int main()

{

	ll n,k,q,m;

	cin>>n>>k>>q;

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

	{

		cin>>m;

		sq[m]++;

	}

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

	{

		if(k-q+sq[i]>0) cout<<"Yes"<<endl;

		else cout<<"No"<<endl;

	}

	return 0;

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

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

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

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

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

		{

			scanf("%lld", &m);

			A[m]++;

		}

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

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.852332072747374x):
#include<iostream>

using namespace std;

typedef long long ll;

const int maxn=1e5+10;

ll sq[maxn];

int main()

{

	ll n,k,q,m;

	cin>>n>>k>>q;

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

	{

		cin>>m;

		sq[m]++;

	}

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

	{

		if(k-q+sq[i]>0) cout<<"Yes"<<endl;

		else cout<<"No"<<endl;

	}

	return 0;

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

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

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

    {

        int x;

        cin>>x;

        a[x]++;

    }

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

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4003971458165652x):
#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <vector>

#include <string>

using namespace std;

typedef long long ll;

const int maxn=2e5+10;

ll n,m,k,d,q;

int a[maxn],level[maxn];

int main()

{

    cin>>n>>k>>q;

    level[0]=n;

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

    {

        int x;

        scanf("%d",&x);

        a[x]++;



    }

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

    {

        if(a[i]>q-k)

            cout<<"Yes"<<endl;

        else

            cout<<"No"<<endl;

    }





    return 0;

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

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

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

{a[i]+=K; a[i]-=Q; }

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

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.119813076653594x):
#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <vector>

#include <string>

using namespace std;

typedef long long ll;

const int maxn=2e5+10;

ll n,m,k,d,q;

int a[maxn],level[maxn];

int main()

{

    cin>>n>>k>>q;

    level[0]=n;

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

    {

        int x;

        scanf("%d",&x);

        a[x]++;



    }

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

    {

        if(a[i]>q-k)

            cout<<"Yes"<<endl;

        else

            cout<<"No"<<endl;

    }





    return 0;

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

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

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

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

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

		{

			scanf("%lld", &m);

			A[m]++;

		}

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

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8475764316374284x):
#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <vector>

#include <string>

using namespace std;

typedef long long ll;

const int maxn=2e5+10;

ll n,m,k,d,q;

int a[maxn],level[maxn];

int main()

{

    cin>>n>>k>>q;

    level[0]=n;

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

    {

        int x;

        scanf("%d",&x);

        a[x]++;



    }

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

    {

        if(a[i]>q-k)

            cout<<"Yes"<<endl;

        else

            cout<<"No"<<endl;

    }





    return 0;

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

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

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

    {

        int x;

        cin>>x;

        a[x]++;

    }

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

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4067054341515501x):
#include <iostream>

using namespace std;



int N, K, Q;

int main(){

	cin>>N>>K>>Q;

	int score[N+1];

	int answer[Q+1];

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

		score[i]=K-Q;

	}

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

		cin>>answer[i];

		score[answer[i]]++;

	}

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

		if(score[i]<=0){

			cout<<"No"<<'\n';

		}else{

			cout<<"Yes"<<'\n';

		}

	}

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

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

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

{a[i]+=K; a[i]-=Q; }

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

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.142876003627731x):
#include <iostream>

using namespace std;



int N, K, Q;

int main(){

	cin>>N>>K>>Q;

	int score[N+1];

	int answer[Q+1];

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

		score[i]=K-Q;

	}

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

		cin>>answer[i];

		score[answer[i]]++;

	}

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

		if(score[i]<=0){

			cout<<"No"<<'\n';

		}else{

			cout<<"Yes"<<'\n';

		}

	}

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

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

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

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

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

		{

			scanf("%lld", &m);

			A[m]++;

		}

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

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8558991027357727x):
#include <iostream>

using namespace std;



int N, K, Q;

int main(){

	cin>>N>>K>>Q;

	int score[N+1];

	int answer[Q+1];

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

		score[i]=K-Q;

	}

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

		cin>>answer[i];

		score[answer[i]]++;

	}

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

		if(score[i]<=0){

			cout<<"No"<<'\n';

		}else{

			cout<<"Yes"<<'\n';

		}

	}

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

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

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

    {

        int x;

        cin>>x;

        a[x]++;

    }

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

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4124454172802792x):


/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



const int mod = 1e9+7;



int n;

int k;

int q;

int cm[N];



int main(){

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

    while(q--){

        int x;

        scanf("%d" , &x);

        cm[0]--;

        cm[x]++;

        cm[x+1]--;

        cm[n+1]++;

    }

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

        cm[i] += cm[i - 1];

        if(k+cm[i] > 0){

            puts("Yes");

        } else {

            puts("No");

        }

    }

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

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

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

{a[i]+=K; a[i]-=Q; }

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

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.163861222549398x):


/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



const int mod = 1e9+7;



int n;

int k;

int q;

int cm[N];



int main(){

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

    while(q--){

        int x;

        scanf("%d" , &x);

        cm[0]--;

        cm[x]++;

        cm[x+1]--;

        cm[n+1]++;

    }

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

        cm[i] += cm[i - 1];

        if(k+cm[i] > 0){

            puts("Yes");

        } else {

            puts("No");

        }

    }

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

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

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

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

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

		{

			scanf("%lld", &m);

			A[m]++;

		}

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

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.863471995595714x):


/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



const int mod = 1e9+7;



int n;

int k;

int q;

int cm[N];



int main(){

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

    while(q--){

        int x;

        scanf("%d" , &x);

        cm[0]--;

        cm[x]++;

        cm[x+1]--;

        cm[n+1]++;

    }

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

        cm[i] += cm[i - 1];

        if(k+cm[i] > 0){

            puts("Yes");

        } else {

            puts("No");

        }

    }

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

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

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

    {

        int x;

        cin>>x;

        a[x]++;

    }

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

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4002069419191816x):
#include <bits/stdc++.h>

using namespace std;

long long n , k , q , arr[100006];

int main() {

	cin>>n>>k>>q;

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

		int x ; cin>>x;

		arr[x]++;

	}

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

		int mi = (q - arr[i]);

		arr[i] = 0;

		arr[i] -= mi;

		arr[i]+=k;

		if(arr[i] <=0)

			cout<<"No"<<endl;

		else cout<<"Yes"<<endl;

	}

	return 0;

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

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

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

{a[i]+=K; a[i]-=Q; }

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

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.119117696486644x):
#include <bits/stdc++.h>

using namespace std;

long long n , k , q , arr[100006];

int main() {

	cin>>n>>k>>q;

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

		int x ; cin>>x;

		arr[x]++;

	}

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

		int mi = (q - arr[i]);

		arr[i] = 0;

		arr[i] -= mi;

		arr[i]+=k;

		if(arr[i] <=0)

			cout<<"No"<<endl;

		else cout<<"Yes"<<endl;

	}

	return 0;

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

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

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

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

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

		{

			scanf("%lld", &m);

			A[m]++;

		}

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

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8473254912245167x):
#include <bits/stdc++.h>

using namespace std;

long long n , k , q , arr[100006];

int main() {

	cin>>n>>k>>q;

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

		int x ; cin>>x;

		arr[x]++;

	}

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

		int mi = (q - arr[i]);

		arr[i] = 0;

		arr[i] -= mi;

		arr[i]+=k;

		if(arr[i] <=0)

			cout<<"No"<<endl;

		else cout<<"Yes"<<endl;

	}

	return 0;

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

using namespace std;

int main()

{

    int n,k,q,a[100010]={0};

    cin>>n>>k>>q;

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

    {

        int x;

        cin>>x;

        a[x]++;

    }

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

        if(k-(q-a[i])<=0)

            cout<<"No"<<endl;

        else

            cout<<"Yes"<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3916190262812134x):
#include <bits/stdc++.h>



using namespace std;



#define el				'\n'

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

#define all(v)          ((v).begin()),((v).end())

#define clr(v, d)		memset(v, d, sizeof(v))



double const EPS = 1e-8, PI = acos(-1);

const int N = 1e5 + 9, M = 2000 + 9, OO = (int) 1e9 + 1;

const long long MOD = 1e9 + 7;



typedef long long ll;



void OUTPUT() {

	cout << fixed << setprecision(12);

	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

}



void INPUT() {

#ifndef ONLINE_JUDGE

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

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

#endif

	cout << fixed << setprecision(0);

//	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

}



ll n, s, q, x, cnt[N];



int main() {

	OUTPUT();

//	INPUT();

	cin >> n >> s >> q;

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

		cin >> x;

		cnt[x - 1]++;

	}

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

		if (s - q + cnt[i] > 0)

			cout << "Yes\n";

		else

			cout << "No\n";

	}



	return 0;



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

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

int main()

{

int N;

long long K,Q;

long long a[500005]={0};

cin>>N;

cin>>K;

cin>>Q;

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

{a[i]+=K; a[i]-=Q; }

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

{

    int op;

    cin>>op;

    a[op]++;

}

for(int j=1;j<=N;j++)

{

    if(a[j]>0)

        printf("Yes\n");

    else

        printf("No\n");

}

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.087720515397111x):
#include <bits/stdc++.h>



using namespace std;



#define el				'\n'

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

#define all(v)          ((v).begin()),((v).end())

#define clr(v, d)		memset(v, d, sizeof(v))



double const EPS = 1e-8, PI = acos(-1);

const int N = 1e5 + 9, M = 2000 + 9, OO = (int) 1e9 + 1;

const long long MOD = 1e9 + 7;



typedef long long ll;



void OUTPUT() {

	cout << fixed << setprecision(12);

	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

}



void INPUT() {

#ifndef ONLINE_JUDGE

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

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

#endif

	cout << fixed << setprecision(0);

//	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

}



ll n, s, q, x, cnt[N];



int main() {

	OUTPUT();

//	INPUT();

	cin >> n >> s >> q;

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

		cin >> x;

		cnt[x - 1]++;

	}

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

		if (s - q + cnt[i] > 0)

			cout << "Yes\n";

		else

			cout << "No\n";

	}



	return 0;



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

#include<cstring>

#include<cstdio>

using namespace std;

long long int A[100005];

int main()

{

	long long int n, k, q, m;

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

	{

		memset(A, 0, sizeof(A));

		for(long long int i = 1; i <= n; i++) A[i] = k - q;

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

		{

			scanf("%lld", &m);

			A[m]++;

		}

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

			if(A[i] <= 0) cout << "No" << endl;

			else cout << "Yes" << endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8359952549576122x):
#include <bits/stdc++.h>



using namespace std;



#define el				'\n'

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

#define all(v)          ((v).begin()),((v).end())

#define clr(v, d)		memset(v, d, sizeof(v))



double const EPS = 1e-8, PI = acos(-1);

const int N = 1e5 + 9, M = 2000 + 9, OO = (int) 1e9 + 1;

const long long MOD = 1e9 + 7;



typedef long long ll;



void OUTPUT() {

	cout << fixed << setprecision(12);

	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

}



void INPUT() {

#ifndef ONLINE_JUDGE

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

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

#endif

	cout << fixed << setprecision(0);

//	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

}



ll n, s, q, x, cnt[N];



int main() {

	OUTPUT();

//	INPUT();

	cin >> n >> s >> q;

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

		cin >> x;

		cnt[x - 1]++;

	}

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

		if (s - q + cnt[i] > 0)

			cout << "Yes\n";

		else

			cout << "No\n";

	}



	return 0;



}