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

using namespace std;

typedef long long ll;

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

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please





int main() {

	cin.tie(0);

	ios::sync_with_stdio(false);





	int N, K, Q;

	cin >> N >> K >> Q;



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		int a;

		cin >> a;

		P[a]++;

	}

	rep1(i, N) {

		if (P[i] > 0) co("Yes");

		else co("No");

	}



	Would you please return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.25113629866955x):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

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

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



const int dm = 1 << 19;

char dn[dm], * di = dn;



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	rep1(i, N) {

		if (P[i] > 0) {

			*di++ = 'Y';

			*di++ = 'e';

			*di++ = 's';

			*di++ = '\n';

		}

		else {

			*di++ = 'N';

			*di++ = 'o';

			*di++ = '\n';

		}

	}

	fwrite(dn, 1, di - dn, stdout);



	Would you please return 0;

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

using namespace std;

typedef long long ll;

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

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please





int main() {

	cin.tie(0);

	ios::sync_with_stdio(false);





	int N, K, Q;

	cin >> N >> K >> Q;



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		int a;

		cin >> a;

		P[a]++;

	}

	rep1(i, N) {

		if (P[i] > 0) co("Yes");

		else co("No");

	}



	Would you please return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.2473154331717335x):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

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

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



const int dm = 1 << 12;

char dn[dm], * di = dn;



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	rep1(i, N) {

		if (P[i] > 0) {

			*di++ = 'Y';

			*di++ = 'e';

			*di++ = 's';

			*di++ = '\n';

		}

		else {

			*di++ = 'N';

			*di++ = 'o';

			*di++ = '\n';

		}

		if (i % 100 == 0) {

			fwrite(dn, 1, di - dn, stdout);

			di = dn;

		}

	}

	fwrite(dn, 1, di - dn, stdout);



	Would you please return 0;

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

#include<cstdio>

#include<string>

using namespace std;

typedef long long ll;

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

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	string S;

	rep1(i, N) {

		if (P[i] > 0) S += "Yes\n";

		else S += "No\n";

	}

	printf("%s", S.c_str());



	Would you please return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.818263066772934x):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

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

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



const int dm = 1 << 12;

char dn[dm], * di = dn;



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	rep1(i, N) {

		if (P[i] > 0) {

			*di++ = 'Y';

			*di++ = 'e';

			*di++ = 's';

			*di++ = '\n';

		}

		else {

			*di++ = 'N';

			*di++ = 'o';

			*di++ = '\n';

		}

		if (i % 100 == 0) {

			fwrite(dn, 1, di - dn, stdout);

			di = dn;

		}

	}

	fwrite(dn, 1, di - dn, stdout);



	Would you please return 0;

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

#include<cstdio>

#include<string>

using namespace std;

typedef long long ll;

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

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	string S;

	rep1(i, N) {

		if (P[i] > 0) S += "Yes\n";

		else S += "No\n";

		if (i % 100 == 0) {

			printf("%s", S.c_str());

			S = "";

		}

	}

	printf("%s", S.c_str());



	Would you please return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.817855250410954x):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

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

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	char C = getcha();

	int A = C - '0';

	while ((C = getcha()) >= '0') A = A * 10 + C - '0';

	return A;

}



const int dm = 1 << 12;

char dn[dm], * di = dn;



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), K = getint(), Q = getint();



	int P[100001];

	rep1(i, N) {

		P[i] = K - Q;

	}

	rep(i, Q) {

		P[getint()]++;

	}

	rep1(i, N) {

		if (P[i] > 0) {

			*di++ = 'Y';

			*di++ = 'e';

			*di++ = 's';

			*di++ = '\n';

		}

		else {

			*di++ = 'N';

			*di++ = 'o';

			*di++ = '\n';

		}

		if (i % 100 == 0) {

			fwrite(dn, 1, di - dn, stdout);

			di = dn;

		}

	}

	fwrite(dn, 1, di - dn, stdout);



	Would you please return 0;

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4073063466170133x):
#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<iostream>

#include<set>

#include<queue>

using namespace std;



int value[555555] = {0};



int main() 

{

	int n, score,m,mark;

	cin >> n >> score >> m;

	int t = m;

	while (t--)

	{

		cin >> mark;

		value[mark-1]++;

	}

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

	{

		if (score > m - value[i])printf("Yes\n");

		else printf("No\n");

	}

	return 0;

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

#define maxn 100005

using namespace std;

int a[maxn];

int main()

{

    int n,k,q;

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

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

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

        int num;

        scanf("%d",&num);

        a[num]++;

    }

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

        if(a[i]>0) puts("Yes");

        else puts("No");

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.6590614215447768x):
#include<stdio.h>

int main()

{

	int n,k,q,i;

    int a;

	int player[100001]={0};



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

    

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

		player[i]=k-q;

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

	{	scanf("%d",&a);

	    player[a]+=1;

	}

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

	{	

		if(player[i]<=0)

	        printf("No\n");

		else

			printf("Yes\n");

	}

return 0;

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.338969323500368x):
#include<stdio.h>

int main()

{

	int n,k,q,i;

    int a;

	int player[100001]={0};



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

    

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

		player[i]=k-q;

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

	{	scanf("%d",&a);

	    player[a]+=1;

	}

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

	{	

		if(player[i]<=0)

	        printf("No\n");

		else

			printf("Yes\n");

	}

return 0;

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

#include<cstring>

#include<cmath>

#include<algorithm>

#include<iostream>

#include<set>

#include<queue>

using namespace std;



int value[555555] = {0};



int main() 

{

	int n, score,m,mark;

	cin >> n >> score >> m;

	int t = m;

	while (t--)

	{

		cin >> mark;

		value[mark-1]++;

	}

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

	{

		if (score > m - value[i])printf("Yes\n");

		else printf("No\n");

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.6620185996623653x):
#include<stdio.h>

int main()

{

	int n,k,q,i;

    int a;

	int player[100001]={0};



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

    

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

		player[i]=k-q;

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

	{	scanf("%d",&a);

	    player[a]+=1;

	}

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

	{	

		if(player[i]<=0)

	        printf("No\n");

		else

			printf("Yes\n");

	}

return 0;

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4081331887476505x):
#include<iostream>

#include<cmath>

#include<algorithm>

using namespace std;

int a[100001] = {0};

int main()

{

	int x,q, k, n;

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

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

	{

		scanf("%d",&x);

		a[x]++;

	}

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

	{

		if (a[i] + k - q > 0)

			printf("Yes\n");

		else

			printf("No\n");

	}

	return 0;

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.38999349807514x):
#include<cstdio>

#include<iostream>

#include<cmath>

#include<cstring>

#include<algorithm>

using namespace std;

int n,k,q;

int a[112000];

int main()

{

	int t;

	cin>>n>>k>>q;

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

	a[i]=k;

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

	{

		cin>>t;

		a[t]++;

	}

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

	{

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

		else cout<<"No"<<endl;

	}

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3896037878484426x):
#include <iostream>

#include <algorithm>

#include <cstring>

#define INF 500050

using namespace std;

int arr[INF],brr[INF];

int main()

{

	int a,b,c;

	cin >> a >> b >> c;

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

		arr[i]=b;

	}

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

		int q;

		cin >> q;

		brr[q]++;

	}

	

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

		arr[i] = arr[i] - c + brr[i];

		if(arr[i] >= 1) cout<<"Yes"<<endl;

		else cout<<"No" <<endl;

	}

	

	return 0;

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.36484349822957x):
#include <cstdio>

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

int a[100000];

int main()

{

	int n,q;

	long long int k;

	int a;

	int t[100000];

	cin >> n >> k >> q;

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

	{

		t[i]=k;

	}

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

	{

		cin >> a;

		t[a]++;

	}

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

	{

		t[i]-=q;

	}

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

	{

		if(t[i]<=0)

		cout << "No" << endl;

		else

		cout << "Yes" << endl;

	}

//	cout << t[8] << " " << t[10]; 

	

	return 0;

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.390106342512393x):
#include <bits/stdc++.h>

using namespace std;	



int n,k,q;

int d[100005];

int main(){

	cin >> n >> k >> q;

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

	{

		int o;

		cin >> o;

		d[o]++;

	}

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

	{

		cout << (k-q+d[i]>0 ? "Yes" : "No") << endl;

		/* code */

	}



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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3749439540314274x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int lint;



int main(){

    lint n, k, q;

    cin >> n >> k >> q;

    lint x = k-q;

    vector <lint> vec;

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

        vec.push_back(x);

    }

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

        lint temp;

        cin >> temp;

        vec[temp]++;

    }

    for(int i=1; i<vec.size(); i++){

        if(vec[i] > 0){

            cout << "Yes" << endl;

        }else{

            cout << "No" << endl;

        }

    }

    return 0;

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3895266416040148x):
#include <cmath>

#include <cstdio>

#include <vector>

#include <iostream>

#include <algorithm>

#include <bits/stdc++.h>

typedef long long int l;

using namespace std;





int main(){

    int n,k,q;

    cin>>n>>k>>q;

    int arr[n],sorgu[q];

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

        cin>>sorgu[i];

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

        arr[i]=k-q;

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

     	arr[sorgu[i]-1]++;

    }

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

        if(arr[i]<=0)

					 cout<<"No"<<endl;

else

cout<<"Yes"<<endl;

    }        

    

    return 0;

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3991678468348956x):
#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

#define ll long long

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

using namespace std;

const int N = 1e5 + 5, M = 3e5 + 5;



int cnt[N];



int main(){

    int n, k, q;

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

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

        int x;

        scanf("%d", &x);

        ++cnt[x];

    }

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

        cnt[i] -= q,

        cnt[i] += k;

        if(cnt[i] > 0)

            puts("Yes");

        else puts("No");

    }

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

#include<cstring>

#include<iostream>

#include<algorithm>

#include<stdlib.h>

#include<string.h>

#include<cmath>

using namespace std;

typedef long long ll;

#define N 100005

int main()

{

    int n,m,i,j,k,l;

    int T,unit;

    int x,z,y,x1,y1;

    int ans1=0,ans2=0;

    int n1,n2;

    int a[100005]={0};

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

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

    {

        scanf("%d",&x);

        a[x]++;

    }

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

    {

        if(m<=T-a[i])

            printf("No\n");

        else

            printf("Yes\n");

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3793901868902092x):
#include<bits/stdc++.h>

#include<iostream>

#include<cmath>

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

typedef long long ll;

using namespace std;

const int inf=0x3f3f3f3f;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=1e5+10;

void io(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);

}

ll a[N];

int main()

{

	io();

	ll n,k,q,b;cin>>n>>k>>q;

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

		cin>>b;

		a[b]++;

	}

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

		if(a[i]+k-q<=0)puts("No");

		else puts("Yes");

	}

	return 0;

}