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: Wrong Answer)

//MADE BY Y_is_sunshine;
//#include <bits/stdc++.h>
//#include <memory.h>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define INF 0x3f3f3f3f
#define MAXN 200005

typedef long long ll;

const ll mod = 1e9 + 7;
const double PI = acos(-1);

using namespace std;

int N, M, K;

struct node {
	ll at;
	ll cost;
	bool operator < (const node a) const {
		return this->at < a.at;
	}
}p[MAXN];

int main(void)
{
#ifdef Sunshine
	freopen("data.txt", "r", stdin);
#endif

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	ll N, M, K;
	cin >> N >> M >> K;
	M *= 2;
	for (int i = 1; i <= N; i++)
		cin >> p[i].at >> p[i].cost;

	sort(p + 1, p + 1 + N);
	ll ans = 0;

	for (int i = 1; i <= N; i++) {
		if (p[i].cost <= 0)
			continue;
		int j = i + 1;
		ll res = p[i].cost / K + (p[i].cost % K == 0ll ? 0 : 1);
		ans += res;
		for (; j <= N; j++) {
			if (p[j].at > p[i].at + M)
				break;
			else
				p[j].cost -= res;
		}
	}

	cout << ans << '\n';
	



#ifdef Sunshine
	freopen("CON", "r", stdin);
	system("pause");
#endif
	return 0;
}