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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.399930667214135x):
#include<bits/stdc++.h>



using namespace std;





typedef long long ll;

const int maxn=2e5+10;

ll a;

//vector<ll> ve;

map<int,int>ma1;

map<int,int>ma2;

int main(){

	ll n,k,q;

	cin>>n>>k>>q;

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

		ma1[i]=k;

	}

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

		cin>>a;

//		for(int j=1;j<=n;j++){

//			if(j!=a[i]) ma[j]--;

//		}

		ma2[a]++;

	}

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

		ma1[i]-=(q-ma2[i]);

	}

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

		if(ma1[i]>0) cout<<"Yes"<<endl;

		else cout<<"No"<<endl;

	}	

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.413908227140872x):
#include<iostream>

using namespace std;

int main(){

    int n,q,a;

    long long int k;

    cin>>n>>k>>q;

    long long int kofn[n];

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

        kofn[i]=k;

    }

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

        cin>>a;

        kofn[a-1]++;

    }

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

        if(kofn[i]-q<=0){

            cout<<"No"<<endl;

            continue;

        }

        cout<<"Yes"<<endl;

    }

    return 0;

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.3917457556549435x):
#include <bits/stdc++.h>



using namespace std;



int main() {

    ios::sync_with_stdio(0) , cin.tie(0) , cout.tie(0);



    int n , k , q;

    cin >> n >> k >> q;



    vector<int> s(n + 1 , 0);

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

        int x;

        cin >> x;

        s[x]++;

    }



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

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

            cout << "Yes\n";

        else

            cout << "No\n";

    }

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4159930274883807x):
#include <bits/stdc++.h>



using namespace std;



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

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

#define pb push_back

#define MP make_pair

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

typedef long long ll;



const ll MOD = 1e9 + 7;

const int OO = (int)1e6;

const int N = (int)1e7;



void init(){

    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}

int a[N];

map<pair<int,int>,int>mp;

int main() {

//	#   ifdef LOCAL_PROJECT

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

//    #   endif

//    int n;

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

//    int start=0,end=1;

//    int ans=0;

//    int cnt =0,pre,cur;

//    scanf("%d",&pre);

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

//        scanf("%d",&cur);

//        if(pre>=cur){cnt++;ans = max(ans,cnt);}

//        else cnt=0;

//        pre= cur;

//    }

//    printf("%d",ans);

    int n,k,q;

    int ans ;

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

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

        mp[{i,k}]=k;

    }

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

        scanf("%d",&ans);

        mp[{ans,k}]++;

    }

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

        if(mp[{i,k}]-q>0)puts("Yes");

        else puts("No");

    }

    return 0;

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.3950112976198499x):
#include<bits/stdc++.h>

using namespace std;

#define LL long long int

#define DL double

#define MLL map<LL,LL>::iterator

#define MSL map<string,L>::iterator

#define MLS map<LL,string>::iterator

#define MSS map<string,string>::iterator

#define MCL map<char,LL>::iterator

#define SL set<LL>::iterator

#define SS set<string>::iterator sl

#define G getline

#define SZ size()

#define IN insert

#define C clear()

#define B begin()

#define F front()

#define T top()

#define E end()

#define EM empty()

#define V vector

#define Q queue

#define DQ deque

#define PQ priority_queue

#define ST stack

#define FI first

#define SE second

#define PI acos(-1)

#define PS push

#define PP pop()

#define PSF push_front

#define PSB push_back

#define PPF pop_front()

#define PPB pop_back()

#define endl "\n"

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

map<LL,LL>v;

int main()

{

    FAST;

    LL n,m,i,j,k,p,q,o,l,s,t,z;

    cin>>n>>m>>s;

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

    {

        cin>>o;

        v[o]++;

    }

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

    {

        if(m+(v[i]-s)<=0)

        {

            cout<<"No"<<endl;

        }

        else

        {

            cout<<"Yes"<<endl;

        }

    }



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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4040905527250893x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long n,i,j,k,l,m,q;

    cin>>n>>k>>q;

    vector<long long>vc(n+10,k);



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

    {

        cin>>l;

        vc[l]++;

    }



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

    {

        if((vc[i]-q)>0)

        {

            cout<<"Yes"<<endl;

        }

        else

        {

            cout<<"No"<<endl;

        }

    }

    return 0;

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4038506087678397x):
#include <iostream>

#include <vector>



using namespace std;



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

{

    int n, k, q;

    cin >> n >> k >> q;

    int player;

    vector<int> scores(n + 1, k);

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

    {

        cin >> player;

        scores[player] += 1;

    }

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

    {

        scores[i] -= q;

        if (scores[i] > 0)

            cout << "Yes" << endl;

        else

            cout << "No" << endl;

    }

    return 0;

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4206683615164666x):
#include<bits/stdc++.h>

#define mp make_pair

#define ff first

#define ss second

#define pb push_back

#define w(x) scanf("%d",&x);while(x--)

#define ws(x) scanf("%d",&x);getchar();while(x--)

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

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

#define pi 2*acos(0.0)

#define endl "\n"

using namespace std;

typedef long long ll;

typedef pair<ll,ll> pll;

typedef vector<int> vi;



const int N=1e6+2;

long long  mod=1e9+9;

int t,j,cs=0,n,a,b,i,k,x;

string s;



int main()

{

    cin>>n>>a>>b;

    k=b-a;

    int ar[n]={};

    for(i=0;i<b;i++){

        cin>>x;

        ar[x-1]++;

    }

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

        if(ar[i]>k)

            cout<<"Yes";

        else

            cout<<"No";

        cout<<endl;

    }

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

using namespace std;

const int MAX=1e5;

int a[MAX];

int main()

{

    memset(a,0,sizeof(a));

    int n,k,q;

    cin>>n>>k>>q;

    int t=q;

    int b;

    while(q--){

        cin>>b;

        a[b-1]++;

    }

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

        if(a[i]>t-k){

            cout<<"Yes"<<endl;

        }

        else{

            cout<<"No"<<endl;

        }

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.406743291823409x):
#include<bits/stdc++.h>

#define mp make_pair

#define ff first

#define ss second

#define pb push_back

#define w(x) scanf("%d",&x);while(x--)

#define ws(x) scanf("%d",&x);getchar();while(x--)

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

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

#define pi 2*acos(0.0)

#define endl "\n"

using namespace std;

typedef long long ll;

typedef pair<ll,ll> pll;

typedef vector<int> vi;



const int N=1e6+2;

long long  mod=1e9+9;

int t,j,cs=0,n,a,b,i,k,x;

string s;



int main()

{

    cin>>n>>a>>b;

    k=b-a;

    int ar[n]={};

    for(i=0;i<b;i++){

        cin>>x;

        ar[x-1]++;

    }

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

        if(ar[i]>k)

            cout<<"Yes";

        else

            cout<<"No";

        cout<<endl;

    }

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.3854287933523413x):
#include<bits/stdc++.h>

#include <ext/rope>



#define min3(a,b,c)   min(a,min(b,c))

#define max3(a,b,c)   max(a,max(b,c))

#define min4(a,b,c,d) min(min(a,b),min(c,d))

#define max4(a,b,c,d) max(max(a,b),max(c,d))



#define count_one(a) __builtin_popcount(a)  // Returns the number of set bits(1) in a number(a). In long long use __builtin_popcountll(a)

#define parity(i)    __builtin_parity(i)  //even parity 0 and odd parity 1

#define blz(a)       __builtin_clz(a) //Returns the number of leading zeroes in a number(a)

#define btz(a)       __builtin_ctz(a) //Returns the number of trailing zeroes in a number(a)

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

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



#define pb   push_back

#define PI   M_PI

#define endl "\n"

#define sc   scanf

#define pf   printf



#define ll  long long

#define ull unsigned long long



#define W(t)            while(t--)

#define rep1(i,n)       for(int i = 0; i < n; i++)

#define rep2(a,b)       for(int i = a; i <= b; i++)

#define rep3(a,b,c)     for(int i = a; i <= b; i+=c)

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

#define repit(it,type)  for(it = type.begin(); it != type.end(); it++)

#define IOS ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

using namespace std;

using namespace __gnu_cxx;



int main()

{

     IOS

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

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

    ll n,k,q,i,x,q1;

    map<ll,ll> freq;



    cin>>n>>k>>q;

    q1 = q;



    W(q)

    {

         cin>>x;

         freq[x]++;

    }

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

    {

         if(freq[i]+k-q1 > 0)

          cout<<"Yes\n";

         else

          cout<<"No\n";

    }



     return 0;

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

using namespace std;

const int MAX=1e5;

int a[MAX];

int main()

{

    memset(a,0,sizeof(a));

    int n,k,q;

    cin>>n>>k>>q;

    int t=q;

    int b;

    while(q--){

        cin>>b;

        a[b-1]++;

    }

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

        if(a[i]>t-k){

            cout<<"Yes"<<endl;

        }

        else{

            cout<<"No"<<endl;

        }

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3718491339294998x):
#include<bits/stdc++.h>

#include <ext/rope>



#define min3(a,b,c)   min(a,min(b,c))

#define max3(a,b,c)   max(a,max(b,c))

#define min4(a,b,c,d) min(min(a,b),min(c,d))

#define max4(a,b,c,d) max(max(a,b),max(c,d))



#define count_one(a) __builtin_popcount(a)  // Returns the number of set bits(1) in a number(a). In long long use __builtin_popcountll(a)

#define parity(i)    __builtin_parity(i)  //even parity 0 and odd parity 1

#define blz(a)       __builtin_clz(a) //Returns the number of leading zeroes in a number(a)

#define btz(a)       __builtin_ctz(a) //Returns the number of trailing zeroes in a number(a)

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

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



#define pb   push_back

#define PI   M_PI

#define endl "\n"

#define sc   scanf

#define pf   printf



#define ll  long long

#define ull unsigned long long



#define W(t)            while(t--)

#define rep1(i,n)       for(int i = 0; i < n; i++)

#define rep2(a,b)       for(int i = a; i <= b; i++)

#define rep3(a,b,c)     for(int i = a; i <= b; i+=c)

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

#define repit(it,type)  for(it = type.begin(); it != type.end(); it++)

#define IOS ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

using namespace std;

using namespace __gnu_cxx;



int main()

{

     IOS

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

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

    ll n,k,q,i,x,q1;

    map<ll,ll> freq;



    cin>>n>>k>>q;

    q1 = q;



    W(q)

    {

         cin>>x;

         freq[x]++;

    }

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

    {

         if(freq[i]+k-q1 > 0)

          cout<<"Yes\n";

         else

          cout<<"No\n";

    }



     return 0;

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<stack>

using namespace std;

#define ll long long

#define mod 1000000007



int a[100005];

int a2[100005];

int main()

{

    int i,t,j,len,temp,N,Q,K;

    while(scanf("%d%d%d",&N,&K,&Q)!=EOF)

    {

        memset(a2,0,sizeof(a2));

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

        {

            scanf("%d",&temp);

            a2[temp]++;

        }

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

        {

            a[i]=K-Q+a2[i]; //不成功的局都得-1分

            if(a[i]<=0)

                printf("No\n");

            else

                printf("Yes\n");

        }

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4175758698634389x):
//std::ios::sync_with_stdio(false);

//std::cin.tie(0);

#include<iostream>

#include<bits/stdc++.h>

#include<cstdio>

#include<string.h>

#include<map>

#include<cmath>

#include<stack>

#include<algorithm>

#include<assert.h>

using namespace std;

#define endl '\n'

#define pb push_back

//#define int long long

#define mp make_pair

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

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

#define mem0(s) memset(s,0,sizeof(s))

#define meminf(s) memset(s,0x3f,sizeof(s))

#define mp_insert(a,b,c) mp.insert(map<pair<int,int>,int>::value_type(make_pair(a,b),c))

double pi=acos(-1);

typedef long long ll;

typedef unsigned long long ull;

inline ll lowbit(ll x){return x&(-x);}

inline ll min(ll a,ll b){return a<b?a:b;}

inline ll max(ll a,ll b){return a>b?a:b;}

const double Rad_to_deg=45.0/atan(1.0);

inline ll floor(const ll&a,const ll&b){return a/b;}

inline ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}

const ll INF=4223372036854775807,max_n=100010,mod=998244353;

inline int cal(int x){int res=0;while(x){x/=10;res++;}return res;}

inline ll ceil(const ll&a,const ll&b){if(a%b==0)return a/b;return a/b+1;};

inline ll pow(ll a,ll b,ll mod){ll res=1;a%=mod;assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

inline int mul(int a,int b){return (ll)a*b%mod;}

int ans[max_n];

int x;

int main()

{

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

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

	{

		cin>>x;

		ans[x]++;

	}

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

	{

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

		{

			cout<<"Yes"<<endl;

		}

		else cout<<"No"<<endl;

	}

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

#include<algorithm>

#include<cstdio>

#include<cmath>

#include<cstring>

using namespace std;

const int MAX=1e5;

int a[MAX];

int main()

{

    memset(a,0,sizeof(a));

    int n,k,q;

    cin>>n>>k>>q;

    int t=q;

    int b;

    while(q--){

        cin>>b;

        a[b-1]++;

    }

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

        if(a[i]>t-k){

            cout<<"Yes"<<endl;

        }

        else{

            cout<<"No"<<endl;

        }

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.403681112073539x):
//std::ios::sync_with_stdio(false);

//std::cin.tie(0);

#include<iostream>

#include<bits/stdc++.h>

#include<cstdio>

#include<string.h>

#include<map>

#include<cmath>

#include<stack>

#include<algorithm>

#include<assert.h>

using namespace std;

#define endl '\n'

#define pb push_back

//#define int long long

#define mp make_pair

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

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

#define mem0(s) memset(s,0,sizeof(s))

#define meminf(s) memset(s,0x3f,sizeof(s))

#define mp_insert(a,b,c) mp.insert(map<pair<int,int>,int>::value_type(make_pair(a,b),c))

double pi=acos(-1);

typedef long long ll;

typedef unsigned long long ull;

inline ll lowbit(ll x){return x&(-x);}

inline ll min(ll a,ll b){return a<b?a:b;}

inline ll max(ll a,ll b){return a>b?a:b;}

const double Rad_to_deg=45.0/atan(1.0);

inline ll floor(const ll&a,const ll&b){return a/b;}

inline ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}

const ll INF=4223372036854775807,max_n=100010,mod=998244353;

inline int cal(int x){int res=0;while(x){x/=10;res++;}return res;}

inline ll ceil(const ll&a,const ll&b){if(a%b==0)return a/b;return a/b+1;};

inline ll pow(ll a,ll b,ll mod){ll res=1;a%=mod;assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

inline int mul(int a,int b){return (ll)a*b%mod;}

int ans[max_n];

int x;

int main()

{

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

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

	{

		cin>>x;

		ans[x]++;

	}

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

	{

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

		{

			cout<<"Yes"<<endl;

		}

		else cout<<"No"<<endl;

	}

}