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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<cmath>
#include<algorithm>
typedef long long ll;
using namespace std;
struct kill {
	ll x;
	ll h;

};
bool cmp(kill x, kill y) {

	return x.x < y.x;


}
int main() {
	ll ans = 0;
	ll n, d, a;
	cin >> n >> d >> a;
	ll dis = 2 * d;
	kill m[200005];
	for (int i = 1; i <= n; i++) cin >> m[i].x >> m[i].h;
	sort(m + 1, m + n + 1, cmp);
	int si = 1, ei;
	while (true) {
		if (si > n) break;
		int pi = 1;
		ll xi = m[si].x + dis;
		while (m[si + pi].x <= xi) { pi++; if ((si + pi) > n) break; }
		ei = si + pi - 1;
		ll k = m[si].h / a;
		if (m[si].h % a != 0) { k++;}
		ans += k;
		bool f1 = false;
		for (int i = si; i <= ei; i++) {
			m[i].h -= k*a;
			if (m[i].h > 0 && f1 == false) { si = i; f1 = true; }

		}
		if (f1 == false) si = ei + 1;


	 }
	
	cout << ans;





	return 0;

}