Problem:
Score : 600 points

Problem Statement

Silver Fox is fighting with N monsters.

The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.

Silver Fox can use bombs to attack the monsters.
Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A.
There is no way other than bombs to decrease the monster's health.

Silver Fox wins when all the monsters' healths become 0 or below.

Find the minimum number of bombs needed to win.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq D \leq 10^9

1 \leq A \leq 10^9

0 \leq X_i \leq 10^9

1 \leq H_i \leq 10^9

X_i are distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N D A
X_1 H_1
:
X_N H_N

Output

Print the minimum number of bombs needed to win.

Sample Input 1

3 3 2
1 2
5 4
9 2

Sample Output 1

2

First, let us use a bomb at the coordinate 4 to decrease the first and second monsters' health by 2.

Then, use a bomb at the coordinate 6 to decrease the second and third monsters' health by 2.

Now, all the monsters' healths are 0.
We cannot make all the monsters' health drop to 0 or below with just one bomb.

Sample Input 2

9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5

Sample Output 2

5

We should use five bombs at the coordinate 5.

Sample Input 3

3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000

Sample Output 3

3000000000

Watch out for overflow.

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val N = sc.nextInt()
    val D = sc.nextLong()
    val A = sc.nextLong()
    val X = mutableListOf<Long>()
    val H = mutableListOf<Long>()
    (1..N)
            .map { Pair(sc.nextLong(), sc.nextLong()) }
            .sortedWith(Comparator { a, b -> if (a.first < b.first) -1 else 1 })
            .forEach { p -> X.add(p.first); H.add(p.second) }
    var ans = 0L
    var count = 0L
    val minus = Array(N + 1) { 0L }
    repeat(N) {
        count -= minus[it]
        val rem = (H[it] - count * A + A - 1) / A
        if (rem > 0) {
            var idx = X.binarySearch(X[it] + D * 2 + 1, it)
            if (idx < 0) {
                idx = idx.inv()
            }
            minus[idx] += rem
            count += rem
            ans += rem
        }
    }
    println(ans)
}