stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 1.3796158469318471x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long 



map<ll,ll>mp,vis;

vector<ll>v;



int main()

{

	ll n, k, i , q, ans, x;



	cin>>n>>k>>q;

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

	{

		cin>>x;

		v.push_back(x);

		mp[x]++;

	}

	sort(v.begin(),v.end());

	v.resize(unique(v.begin(),v.end())-v.begin());

	ll sum=0;

	for(i=0; i<v.size(); i++)

	{

		  sum+=mp[v[i]];

	}

	ll baki=0;

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

	{

		baki=sum;

		if(mp[i]!=0)

		{

		 baki=sum-mp[i];

		}

		if((k-baki)>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: 5.043837225019173x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long 



map<ll,ll>mp,vis;

vector<ll>v;



int main()

{

	ll n, k, i , q, ans, x;



	cin>>n>>k>>q;

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

	{

		cin>>x;

		v.push_back(x);

		mp[x]++;

	}

	sort(v.begin(),v.end());

	v.resize(unique(v.begin(),v.end())-v.begin());

	ll sum=0;

	for(i=0; i<v.size(); i++)

	{

		  sum+=mp[v[i]];

	}

	ll baki=0;

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

	{

		baki=sum;

		if(mp[i]!=0)

		{

		 baki=sum-mp[i];

		}

		if((k-baki)>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: 1.8201591820715344x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long 



map<ll,ll>mp,vis;

vector<ll>v;



int main()

{

	ll n, k, i , q, ans, x;



	cin>>n>>k>>q;

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

	{

		cin>>x;

		v.push_back(x);

		mp[x]++;

	}

	sort(v.begin(),v.end());

	v.resize(unique(v.begin(),v.end())-v.begin());

	ll sum=0;

	for(i=0; i<v.size(); i++)

	{

		  sum+=mp[v[i]];

	}

	ll baki=0;

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

	{

		baki=sum;

		if(mp[i]!=0)

		{

		 baki=sum-mp[i];

		}

		if((k-baki)>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: 1.4034283914240413x):
#include <bits/stdc++.h>

using namespace std;

int n , k , q , p[100005] , a;

int main(){

	cin >> n >> k >> q;

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

		cin >> a;

		p[a]++;

	}

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

		if(p[i] - q + k <= 0)cout << "No" << endl;

		else cout << "Yes" << 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: 5.130895226418085x):
#include <bits/stdc++.h>

using namespace std;

int n , k , q , p[100005] , a;

int main(){

	cin >> n >> k >> q;

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

		cin >> a;

		p[a]++;

	}

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

		if(p[i] - q + k <= 0)cout << "No" << endl;

		else cout << "Yes" << 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: 1.8515756242661818x):
#include <bits/stdc++.h>

using namespace std;

int n , k , q , p[100005] , a;

int main(){

	cin >> n >> k >> q;

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

		cin >> a;

		p[a]++;

	}

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

		if(p[i] - q + k <= 0)cout << "No" << endl;

		else cout << "Yes" << 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: 1.4219879456792455x):
#include <bits/stdc++.h>

using namespace std;

int a[100005], freq[100005], health[100005];



int main() {

    int n, k, q;

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

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

        scanf("%d", &a[i]);

        freq[a[i]]++;

    }

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

        health[i] = k - (q - freq[i]);

        if(health[i] > 0) {

            printf("Yes\n");

        } else {

            printf("No\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: 5.198748441383936x):
#include <bits/stdc++.h>

using namespace std;

int a[100005], freq[100005], health[100005];



int main() {

    int n, k, q;

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

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

        scanf("%d", &a[i]);

        freq[a[i]]++;

    }

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

        health[i] = k - (q - freq[i]);

        if(health[i] > 0) {

            printf("Yes\n");

        } else {

            printf("No\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: 1.8760616746170036x):
#include <bits/stdc++.h>

using namespace std;

int a[100005], freq[100005], health[100005];



int main() {

    int n, k, q;

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

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

        scanf("%d", &a[i]);

        freq[a[i]]++;

    }

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

        health[i] = k - (q - freq[i]);

        if(health[i] > 0) {

            printf("Yes\n");

        } else {

            printf("No\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: 1.3718438022151085x):
////////////////////////// Author

////////////////////////// Nasim Hossain Rabbi

////////////////////////// JU-CSE28

////////////////////////// CF - imnasim3.1415

////////////////////////// UVA - imnasim3.1415

////////////////////////// Mail - imnasim3.1415@gmail.com

#include<bits/stdc++.h>

using namespace std;

#define sz size()

#define mk make_pair

#define clr clear()

#define B begin()

#define E end()

#define EM empty()

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

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

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

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

#define pb push_back

string en="\n";

string t="hello";

#define sqr(a) ((a) * (a))

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

#define pi 3.1415926535897932384626433832795

#define vi vector<int>

#define vl vector<long long>

#define vs vector<string>

#define vp(vi,x) cin>>x; vi.pb(x);

#define LB(a,x) (lower_bound(all(a),x)-a.B)

#define UB(a,x) (upper_bound(all(a),x)-a.B)

typedef long long LL;

typedef unsigned long long ULL;

bool isVowel(char ch){ ch=toupper(ch); if(ch=='A'||ch=='U'||ch=='I'||ch=='O'||ch=='E') return true; return false;}

int toInt(string s){int sm;stringstream ss(s);ss>>sm;return sm;}

template<typename T>inline T gcd(T a,T b){if(a<0)return gcd(-a,b);if(b<0)return gcd(a,-b);return (b==0)?a:gcd(b,a%b);}

template<typename T>inline T lcm(T a,T b) {if(a<0)return lcm(-a,b);if(b<0)return lcm(a,-b);return a*(b/gcd(a,b));}

int main()

{

//    fast;

    LL i,j,k,n,m,l,s=0,x,y,tc=1,c=0;

    while(cin>>n>>k>>m)

    {

        vi a;

        map<int,int>mp;

        rep(i,m)

        {

            cin>>x;

            mp[x]++;

        }

        Rep(i,n)

        {

            y=k-(m-mp[i]);

            if(y>0) cout<<"Yes"<<en;

            else cout<<"No"<<en;

        }



    }

}
// 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: 5.015422845361257x):
////////////////////////// Author

////////////////////////// Nasim Hossain Rabbi

////////////////////////// JU-CSE28

////////////////////////// CF - imnasim3.1415

////////////////////////// UVA - imnasim3.1415

////////////////////////// Mail - imnasim3.1415@gmail.com

#include<bits/stdc++.h>

using namespace std;

#define sz size()

#define mk make_pair

#define clr clear()

#define B begin()

#define E end()

#define EM empty()

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

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

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

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

#define pb push_back

string en="\n";

string t="hello";

#define sqr(a) ((a) * (a))

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

#define pi 3.1415926535897932384626433832795

#define vi vector<int>

#define vl vector<long long>

#define vs vector<string>

#define vp(vi,x) cin>>x; vi.pb(x);

#define LB(a,x) (lower_bound(all(a),x)-a.B)

#define UB(a,x) (upper_bound(all(a),x)-a.B)

typedef long long LL;

typedef unsigned long long ULL;

bool isVowel(char ch){ ch=toupper(ch); if(ch=='A'||ch=='U'||ch=='I'||ch=='O'||ch=='E') return true; return false;}

int toInt(string s){int sm;stringstream ss(s);ss>>sm;return sm;}

template<typename T>inline T gcd(T a,T b){if(a<0)return gcd(-a,b);if(b<0)return gcd(a,-b);return (b==0)?a:gcd(b,a%b);}

template<typename T>inline T lcm(T a,T b) {if(a<0)return lcm(-a,b);if(b<0)return lcm(a,-b);return a*(b/gcd(a,b));}

int main()

{

//    fast;

    LL i,j,k,n,m,l,s=0,x,y,tc=1,c=0;

    while(cin>>n>>k>>m)

    {

        vi a;

        map<int,int>mp;

        rep(i,m)

        {

            cin>>x;

            mp[x]++;

        }

        Rep(i,n)

        {

            y=k-(m-mp[i]);

            if(y>0) cout<<"Yes"<<en;

            else cout<<"No"<<en;

        }



    }

}
// 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: 1.8099053432321919x):
////////////////////////// Author

////////////////////////// Nasim Hossain Rabbi

////////////////////////// JU-CSE28

////////////////////////// CF - imnasim3.1415

////////////////////////// UVA - imnasim3.1415

////////////////////////// Mail - imnasim3.1415@gmail.com

#include<bits/stdc++.h>

using namespace std;

#define sz size()

#define mk make_pair

#define clr clear()

#define B begin()

#define E end()

#define EM empty()

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

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

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

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

#define pb push_back

string en="\n";

string t="hello";

#define sqr(a) ((a) * (a))

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

#define pi 3.1415926535897932384626433832795

#define vi vector<int>

#define vl vector<long long>

#define vs vector<string>

#define vp(vi,x) cin>>x; vi.pb(x);

#define LB(a,x) (lower_bound(all(a),x)-a.B)

#define UB(a,x) (upper_bound(all(a),x)-a.B)

typedef long long LL;

typedef unsigned long long ULL;

bool isVowel(char ch){ ch=toupper(ch); if(ch=='A'||ch=='U'||ch=='I'||ch=='O'||ch=='E') return true; return false;}

int toInt(string s){int sm;stringstream ss(s);ss>>sm;return sm;}

template<typename T>inline T gcd(T a,T b){if(a<0)return gcd(-a,b);if(b<0)return gcd(a,-b);return (b==0)?a:gcd(b,a%b);}

template<typename T>inline T lcm(T a,T b) {if(a<0)return lcm(-a,b);if(b<0)return lcm(a,-b);return a*(b/gcd(a,b));}

int main()

{

//    fast;

    LL i,j,k,n,m,l,s=0,x,y,tc=1,c=0;

    while(cin>>n>>k>>m)

    {

        vi a;

        map<int,int>mp;

        rep(i,m)

        {

            cin>>x;

            mp[x]++;

        }

        Rep(i,n)

        {

            y=k-(m-mp[i]);

            if(y>0) cout<<"Yes"<<en;

            else cout<<"No"<<en;

        }



    }

}
// 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: 1.4113266083947278x):
#include <bits/stdc++.h>

#define ll long long

#define N 100010

using namespace std;

ll n, k, q, point[N];

int main()

{

    cin>>n>>k>>q;

    for(ll i=1; i<=n; i++) point[i]=k;

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

    {

        ll x;cin>>x;

        point[x]++;

    }

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

    {

        if (point[i] - q > 0) puts("Yes");

        else puts("No");

    }

    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: 5.159770888332684x):
#include <bits/stdc++.h>

#define ll long long

#define N 100010

using namespace std;

ll n, k, q, point[N];

int main()

{

    cin>>n>>k>>q;

    for(ll i=1; i<=n; i++) point[i]=k;

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

    {

        ll x;cin>>x;

        point[x]++;

    }

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

    {

        if (point[i] - q > 0) puts("Yes");

        else puts("No");

    }

    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: 1.8619959250862688x):
#include <bits/stdc++.h>

#define ll long long

#define N 100010

using namespace std;

ll n, k, q, point[N];

int main()

{

    cin>>n>>k>>q;

    for(ll i=1; i<=n; i++) point[i]=k;

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

    {

        ll x;cin>>x;

        point[x]++;

    }

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

    {

        if (point[i] - q > 0) puts("Yes");

        else puts("No");

    }

    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: 1.3786480035845863x):
/******************************************

* AUTHOR : RAJAGOPALAN *

* NICK : ARNO *

* INSTITUTION : VIT *

******************************************/

#include <bits/stdc++.h>

#include <ext/pb_ds/detail/standard_policies.hpp>

#define ll long long

#define ull unsigned long long

#define pb push_back

#define mp make_pair

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

#define alli(a, n, k) (a+k),(a+n+k)

#define FP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)

#define FS(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)

#define IT(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)

#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>

#define eps 1e-6

#define pi 3.141592653589793

using namespace std;

using namespace __gnu_pbds;



template<class T> inline T gcd(T x, T y) { if (!y) return x; return gcd(y, x%y);}

typedef vector<int> VII;

typedef vector<ll> VLL;

typedef pair<int, int> PII;

typedef vector< pair<int, int> > VPII;

typedef vector< pair<int, PII > > VPPI;

const int MOD = 1e9 + 7;

const int INF = 1e9;



inline ll modulo(ll a, ll m)

{

    return (a % m + m) % m;

}



inline ll modInverse(ll a, ll m)

{

    assert(__gcd(a, m) == 1);

    ll m0 = m;

    ll y = 0, x = 1;



    if (m == 1)

        return 0;



    while (a > 1)

    {

        ll q = a / m;

        ll t = m;



        m = a % m, a = t;

        t = y;



        y = x - q * y;

        x = t;

    }



    if (x < 0)

        x += m0;



    return x;

}



inline ll modPow(ll x, ll y, ll m)

{ //x^y % m

    if (y == 0)

        return 1LL;

    else if (y == 1)

        return x;

    else

    {

        ll ans = modPow(x, y / 2, m) % m;

        if (y & 1)

        {

            return (((ans * ans) % m) * x) % m;

        }

        else

        {

            return (ans * ans) % m;

        }

    }

}



int main(int argc, char* argv[])

{

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    int n,k,q;

    cin>>n>>k>>q;

    VII s(n+1,k);

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

    {

        int x;

        cin>>x;

        s[x]++;

    }

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

    {

        if(s[i]-q>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: 5.040298816618198x):
/******************************************

* AUTHOR : RAJAGOPALAN *

* NICK : ARNO *

* INSTITUTION : VIT *

******************************************/

#include <bits/stdc++.h>

#include <ext/pb_ds/detail/standard_policies.hpp>

#define ll long long

#define ull unsigned long long

#define pb push_back

#define mp make_pair

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

#define alli(a, n, k) (a+k),(a+n+k)

#define FP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)

#define FS(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)

#define IT(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)

#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>

#define eps 1e-6

#define pi 3.141592653589793

using namespace std;

using namespace __gnu_pbds;



template<class T> inline T gcd(T x, T y) { if (!y) return x; return gcd(y, x%y);}

typedef vector<int> VII;

typedef vector<ll> VLL;

typedef pair<int, int> PII;

typedef vector< pair<int, int> > VPII;

typedef vector< pair<int, PII > > VPPI;

const int MOD = 1e9 + 7;

const int INF = 1e9;



inline ll modulo(ll a, ll m)

{

    return (a % m + m) % m;

}



inline ll modInverse(ll a, ll m)

{

    assert(__gcd(a, m) == 1);

    ll m0 = m;

    ll y = 0, x = 1;



    if (m == 1)

        return 0;



    while (a > 1)

    {

        ll q = a / m;

        ll t = m;



        m = a % m, a = t;

        t = y;



        y = x - q * y;

        x = t;

    }



    if (x < 0)

        x += m0;



    return x;

}



inline ll modPow(ll x, ll y, ll m)

{ //x^y % m

    if (y == 0)

        return 1LL;

    else if (y == 1)

        return x;

    else

    {

        ll ans = modPow(x, y / 2, m) % m;

        if (y & 1)

        {

            return (((ans * ans) % m) * x) % m;

        }

        else

        {

            return (ans * ans) % m;

        }

    }

}



int main(int argc, char* argv[])

{

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    int n,k,q;

    cin>>n>>k>>q;

    VII s(n+1,k);

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

    {

        int x;

        cin>>x;

        s[x]++;

    }

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

    {

        if(s[i]-q>0)

        cout<<"Yes\n";



        else cout<<"No\n";

    }





    return 0;

}