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++14 (GCC 5.4.1) (judge verdict: Accepted)

//#pragma GCC optimize("Ofast")

#pragma GCC optimize("O3")
//#pragma GCC target("avx,avx2,fma")
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstdlib>
#include<set>
#include<vector>
#include<sstream>
#include<queue>
#include <iomanip>
#include<unordered_set>
#include <stack>
#include <unordered_map>
#include <map>
constexpr auto INF = 9223372036854775807;
typedef long long int ll;
typedef unsigned long long int ull;
typedef unsigned long int ul;
#define f(i,a,b) for(ll i=(ll)a;i<(ll)b;i+=1)
#define rf(i,a,b) for(ll i=(ll)a;i>=(ll)b;i-=1)
#define endl '\n'
#define N 1000000007 // prime modulo value 
#define M 998244353
#define all(x) x.begin(),x.end()
using namespace std;
// From Geeksforgeeks
inline ll gcd(ll a, ll b)
{
	if (b == 0)
		return a;
	return gcd(b, a % b);
}
// Power function took from geeksforgeeks
ll power(ll x, ll y)
{
	ll res = 1;

	x = x%N;
	if (x == 0) return 0;
	while (y > 0)
	{
		if (y & 1)
			res = (res * x)%N;
		y = y >> 1;
		x = (x * x)%N;
	}
	return res;
}
double dist(ll x1,ll y1, ll x2,ll y2)
{
	double ans = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
	return ans;
}
int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	ll n, k, q;
	cin >> n >> k >> q;
	vector<ll> arr(n, 0);
	f(i, 0, q)
	{
		ll temp;
		cin >> temp;
		arr[temp - 1] += 1;
	}
	ll sum = 0;
	f(i, 0, n)
	{
		sum += arr[i];
	}
	f(i, 0, n)
	{
		if (sum - arr[i] >= k)
			cout << "No" << endl;
		else
			cout << "Yes" << endl;;
	}
	return 0;
}