Problem:
Score: 300 points

Problem Statement

Takahashi has decided to hold fastest-finger-fast quiz games. Kizahashi, who is in charge of making the scoreboard, is struggling to write the program that manages the players' scores in a game, which proceeds as follows.

A game is played by N players, numbered 1 to N. At the beginning of a game, each player has K points.

When a player correctly answers a question, each of the other N-1 players receives minus one (-1) point. There is no other factor that affects the players' scores.

At the end of a game, the players with 0 points or lower are eliminated, and the remaining players survive.

In the last game, the players gave a total of Q correct answers, the i-th of which was given by Player A_i.
For Kizahashi, write a program that determines whether each of the N players survived this game.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq K \leq 10^9

1 \leq Q \leq 10^5

1 \leq A_i \leq N\ (1 \leq i \leq Q)

Input

Input is given from Standard Input in the following format:

N K Q
A_1
A_2
.
.
.
A_Q

Output

Print N lines. The i-th line should contain Yes if Player i survived the game, and No otherwise.

Sample Input 1

6 3 4
3
1
3
2

Sample Output 1

No
No
Yes
No
No
No

In the beginning, the players' scores are (3, 3, 3, 3, 3, 3).

Player 3 correctly answers a question. The players' scores are now (2, 2, 3, 2, 2, 2).

Player 1 correctly answers a question. The players' scores are now (2, 1, 2, 1, 1, 1).

Player 3 correctly answers a question. The players' scores are now (1, 0, 2, 0, 0, 0).

Player 2 correctly answers a question. The players' scores are now (0, 0, 1, -1, -1, -1).

Players 1, 2, 4, 5 and 6, who have 0 points or lower, are eliminated, and Player 3 survives this game.

Sample Input 2

6 5 4
3
1
3
2

Sample Output 2

Yes
Yes
Yes
Yes
Yes
Yes

Sample Input 3

10 13 15
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9

Sample Output 3

No
No
No
No
Yes
No
No
No
Yes
No

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = (0); i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define show(x) cout << #x << " = " << (x) << endl
#define show2(x,y) cout << #x << " = " << (x) << ", " << #y << " = " << (y) << endl
#define show3(x,y,z) cout << #x << " = " << (x) << ", " << #y << " = " << (y) <<  ", " << #z << " = " << (z) << endl
#define showv(v) rep(i,v.size()) printf("%d%c", v[i], i==v.size()-1?'\n':' ')
#define showv2(v) rep(j,v.size()) showv(v[j])
#define showt(t,n) rep(i,n) printf("%d%c", t[i], i==n-1?'\n':' ')
#define showt2(t,r,c) rep(j,r) showt(t[j],c)
#define showvp(p) rep(i,p.size()) printf("%d %d\n", p[i].first, p[i].second)
#define printv(v) rep(i,v.size()) printf("%d\n", v[i])
#define printt(t,n) rep(i,n) printf("%d\n", t[i])
#define incl(v,x) (find(all(v),x)!=v.end())
#define incls(s,c) (s.find(c)!=string::npos)
#define lb(a,x) distance((a).begin(),lower_bound(all(a),(x)))
#define ub(a,x) distance((a).begin(),upper_bound(all(a),(x)))
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define bit(n,k) ((n>>k)&1) // nのk bit目
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define uni(x) x.erase(unique(all(x)),x.end())
#define SP << " " <<
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
using vb = vector<bool>;
using vvb = vector<vb>;
using P = pair<int, int>;
using T = tuple<int, int, int>;
using vp = vector<P>;
using vt = vector<T>;
const int mod = 1000000007;
const double EPS = 1e-9;
//const long double EPS = 1e-14;
const int INF = (1<<30)-1;
const ll LINF = (1LL<<62)-1;
#define dame { puts("No"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
inline int in() { int x; cin >> x; return x;}
inline ll lin() { ll x; cin >> x; return x;}
inline char chin() { char x; cin >> x; return x;}
inline string stin() { string x; cin >> x; return x;}
inline double din() { double x; cin >> x; return x;}
//template<class T = int> inline T in() { T x; cin >> x; return (x);}
template<typename T>inline ll suma(const vector<T>& a) { ll res(0); for (auto&& x : a) res += x; return res;}
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
char itoa(int n) { return n + '0';}
ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
const int dx[4] = {1, 0,-1, 0};
const int dy[4] = {0, 1, 0,-1};

int main () {
  int n, k, q;
  cin >> n >> k >> q;
  vi cnt(n,k-q);
  rep(i,q) {
    int a = in()-1;
    cnt[a]++;
  }
  rep(i,n) {
    if (cnt[i] > 0) puts("Yes");
    else puts("No");
  }
  return 0;
}