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)

#include <iostream>

#include <vector>
#include <map>
#include <algorithm>
#include <fstream>
#include<cstdio>
#include<iomanip>
#include<stack>
#include<queue>
#include<string>
#include <cstdlib>
#include <typeinfo>
#include <math.h>
#include <list>

#define REP(i, n) for(int i=0;i<n;i++)
#define REP2(i, s, n) for(int i=s;i<n;i++)
#define REP_1(i, n) for(int i=1;i<n+1;i++)
#define bitSearch(bit, n) for(int bit = 0; bit < (1 << n); bit++)
#define sz(x) int(x.size())
using namespace std;

template<class T>
void print(const T &value) {
    std::cout << value << std::endl;
}

void yesno(bool a) { if (a)cout << "yes" << endl; else cout << "no" << endl; }

void YesNo(bool a) { if (a)cout << "Yes" << endl; else cout << "No" << endl; }

void YESNO(bool a) { if (a)cout << "YES" << endl; else cout << "NO" << endl; }

typedef long long ll;
typedef unsigned long ul;
typedef long double ld;

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

//template<class T>
//inline bool chmax(T &a, T b) {
//    if (a < b) {
//        a = b;
//        return true;
//    }
//    return false;
//}


ll INF = 1000001000;
ll mod = 1000000007;//10^9+7
ll power(ll x, ll n, ll mod) {//x^nをmodで割った余り
    ll res = 1;
    if (n > 0) {
        res = power(x, n / 2, mod);
        if (n % 2 == 0) res = (res * res) % mod;
        else res = (((res * res) % mod) * x) % mod;
    }
    return res;
}

int dx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};

struct Edge {
    int to, cost;

    Edge(int to, int cost) : to(to), cost(cost) {};
};

using Graph = vector<vector<int>>;//隣接リスト:G[i]にはiと隣接する頂点が入るよ。
using GraphW = vector<vector<Edge>>;//重み付きの隣接リスト
using P = pair<int, int>;//BFSで利用。queueに入れる。
using lP = pair<ll, ll>;
using PP = pair<int, P>; //dijkstraで利用,priority_queueに入れる。
using p_queue = priority_queue<int, vector<int>, greater<int>>;//小さい順

class UnionFindTree {

public:
    UnionFindTree(int size) : memberSize(size) {
        par.resize(size * sizeof(ll));
        rnk.resize(size * sizeof(ll));
        diff_weight.resize(size * sizeof(ll));
        REP(i, size) {
            par[i] = i;
            rnk[i] = 0;
            diff_weight[i] = 0;
        }
    }

    int memberSize;

    vector<int> par;
    vector<int> rnk;
    vector<ll> diff_weight;

//n個のnodeで初期化する関数
    void init(ll n) {
        REP(i, n) {
            par[i] = i;
            rnk[i] = 1;
        }
    }

//木の根を求める。この際，親は根に更新される。
    int find(ll x) {
        if (par[x] == x) { return x; }
        else {
            int r = find(par[x]);
            diff_weight[x] += diff_weight[par[x]];
            return par[x] = r;
        }
    }

//xとyが同じ集合に属するか判定
    bool same(ll x, ll y) {
        return find(x) == find(y);
    }

//xとyの属する集合を融合
    void unite(ll x, ll y) {
        x = find(x);
        y = find(y);
        if (x == y) { return; }

        if (rnk[x] < rnk[y]) {
            par[x] = y;
        } else {
            par[y] = x;
            if (rnk[x] == rnk[y]) { rnk[x]++; }
        }
    }

    //xの重み
    ll weight(ll x) {
        find(x);
        return diff_weight[x];
    }

    ll diff(ll x, ll y) {
        return weight(y) - weight(x);
    }

    //重みごと更新。weight(y)-weight(x)=wとなるようにする
    void merge(int x, int y, int w) {
        w += weight(x);
        w -= weight(y);
        x = find(x);
        y = find(y);
        if (x == y) { return; }
        if (rnk[x] < rnk[y]) {
            par[x] = y;
            diff_weight[x] = -w;
        } else {
            par[y] = x;
            diff_weight[y] = w;
            if (rnk[x] == rnk[y]) { rnk[x]++; }
        }
    }
};

priority_queue<P, vector<P>, greater<P>> que;

struct Coordinate {
    int x, y;

    Coordinate(int y, int x) : y(y), x(x) {};
};

ll gcd(ll a, ll b) {
    if (a < b) {
        ll tmp = a;
        a = b;
        b = tmp;
    }
    ll r = a % b;
    while (r != 0) {
        a = b;
        b = r;
        r = a % b;
    }
    return b;
}

bool chmax(ld &a, ld b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

//番号ズレ注意！！
int main() {
    ll N, K, Q;
    cin >> N >> K >> Q;
    ll point[N];
    fill(point, point + N, K - Q);
    REP(i, Q) {
        int A;
        cin >> A;
        A--;
        point[A] += 1;
    }
    REP(i,N){
        YesNo(point[i]>0);
    }
}