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 PHP (7.4.4) (judge verdict: Time Limit Exceeded)

<?php

[$N, $D, $A] = fscanf(STDIN, "%d%d%d");
$mons = [];
for($i=0;$i<$N;$i++){
    [$x, $h] = fscanf(STDIN, "%d%d");
    $mons[$x] = $h;
}
ksort($mons);
$k = [[-1,0]];
$know = 0;
$kmax = 0;
$ans = 0;
foreach($mons as $x => $h){
    for($j=$know;$j<=$kmax;$j++){
        if($x> $k[$j][0]) {
            if($know != $kmax){
                $know++;
            }
        }elseif($x <= $k[$j][0]){
            $h-= $k[$j][1];
            if($h<=0){
                continue 2;
            }else{
                break;
            }
        }
    }
    $hiku = intdiv($h + $A - 1, $A);

    $ans += $hiku;
    for($j=$know;$j<=$kmax;$j++){
        $k[$j][1] += $hiku*$A;
    }
    $kmax++;
    $k[$kmax] = [$x + $D +$D, $hiku * $A];
}
echo $ans;