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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

import java.lang.Math.abs
import java.math.BigInteger

fun read() = readLine()!!

operator fun BigInteger.dec(): BigInteger = this.subtract(BigInteger.ONE)

// 階乗 (n!)
tailrec fun fact(n: BigInteger, ans: BigInteger = BigInteger.ONE) : BigInteger {
    if(n.compareTo(BigInteger.ONE) < 1) return ans
    return fact(n.dec(), ans * n)
}

// 最大公約数
tailrec fun gcd(a: Int, b: Int): Int {
    if(b == 0) return a
    return gcd(b, a % b)
}

// フィボナッチ数
//
tailrec fun fib(n: BigInteger, a: BigInteger = BigInteger.ZERO, b: BigInteger = BigInteger.ONE): BigInteger {
    if(n == BigInteger.ONE) return a
    return fib(n.dec(), b, a + b)
}

// 累乗(バイナリ法)
// a^n
tailrec fun pow(a: BigInteger, n: Int, ans: BigInteger = BigInteger.ONE): BigInteger {
    if(n == 0) return ans
    return pow(a * a, n.ushr(1), if(n.and(1) == 1) ans * a else ans)
}

tailrec fun newton(
    f: (Double) -> Double,
    df: (Double) -> Double,
    x: Double,
    eps: Double = Float.MIN_VALUE.toDouble()
): Double {
    val xnew = x - f(x) / df(x)
    if(abs(xnew - x) < eps) return xnew
    return newton(f, df, xnew, eps)
}

tailrec fun iteration(
    r: (Double) -> Double,
    x: Double,
    eps: Double = Float.MIN_VALUE.toDouble()
): Double {
    val xnew = r(x)
    if(abs(xnew - x) < eps) return xnew
    return iteration(r, xnew, eps)
}

// 二分探索 (木構造使ってない)
fun binarySearch(list: List<Int>, target: Int): Int {
    if(target < list.first() || list.last() < target) return -1

    var first = 0
    var last = list.size
    while (first + 1 != last) {
        val center = first + (last - first) / 2
        val centerVal = list[center]

        if(centerVal == target) return centerVal

        if(centerVal < target) first = center
        else last = center
    }
    return -1
}

fun main(vararg args: String) {
    val (n, k, q) = read().split(" ").map(String::toInt)
    val a = mutableListOf<Int>()
    for(i in 0 until q) a.add(read().toInt())

    val tmp = k - q
    val s = buildString {
        for (i in 0 until n) {
            val c = a.count { it == i + 1 }
            appendln(if(tmp + c <= 0) "No" else "Yes")
        }
    }
    print(s)
}