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 C (GCC 5.4.1) (judge verdict: Accepted)

// ABC 153-F
// 2020.1.26 bal4u

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef long long ll;

#define gc() getchar_unlocked()

int in() {   // 非負整数の入力
	int n = 0; int c;
	c = gc();
	do n = 10 * n + (c & 0xf); while ((c = gc()) >= '0');
	return n;
}

#define MAX 200005
typedef struct { int X, H; } T;
T a[MAX]; int N;

void Usort(T *a, int n) {
	int i, k;
#define B 8
#define BMSK 255 // (1<<8)-1
	int tt1[1 << B], tt2[1 << B];
	T *tmp = malloc(n * sizeof(T));
	
	for (k = 0; k < 4; ++k) {
		memset(tt1, 0, sizeof(tt1)), memset(tt2, 0, sizeof(tt2));
		for (i = 0; i < n; ++i)	tt1[a[i].X >> k * B & BMSK]++;
		for (i = 1; i < (1 << B); ++i) tt1[i] += tt1[i-1];
		i = n; while (i--) tmp[--tt1[a[i].X >> k * B & BMSK]] = a[i];
		k++;
		for (i = 0; i < n; ++i) tt2[tmp[i].X >> k * B & BMSK]++;
		for (i = 1; i < (1 << B); ++i) tt2[i] += tt2[i-1];
		i = n; while (i--) a[--tt2[tmp[i].X >> k * B & BMSK]] = tmp[i];
	}
	free(tmp);
}

// 見つからなければ、一つ小さい要素を返す
int bs(int l, int x) {
	int m, r = N+1;

    while (l < r) {
        m = (l+r) >> 1;
        if (a[m].X <= x) l = m + 1; else r = m;
    }
	return l-1;
}

ll imos[MAX];

int main()
{
	int i, j, A, D;
	ll h, t, ans;
	
	N = in(), D = in(), A = in();
	for (i = 1; i <= N; ++i) a[i].X = in(), a[i].H = in();
	Usort(a+1, N);

	if ((ll)D*2 > a[N].X) D = a[N].X; else D *= 2;
	
	ans = 0;
	for (i = 1; i <= N; ++i) {
		imos[i] += imos[i-1];
		h = a[i].H - imos[i];
		if (h > 0) {
			t = (h-1)/A + 1;
			ans += t, t *= A;
			j = bs(i, a[i].X + D);
			imos[i] += t, imos[j+1] -= t;
		}
	}
	printf("%lld\n", ans);
	return 0;
}