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: Accepted)

#pragma target("avx")
#include<bits/stdc++.h>
using ll = int_fast64_t;
using P = std::pair<ll,ll>;
using V = std::vector<ll>;
template<typename T> using pq = std::priority_queue<T>;
template<typename T> using rpq = std::priority_queue<T, std::vector<T>, std::greater<T>>;

#define rep(i,b,e) for(ll i=b; i<e; i++)
#define be(vec) vec.begin(),vec.end()
#define print(vec) printf("[ ");for(auto &i:vec)printf("%ld ",i);puts("]");
#define scan(vec) for(auto &i:vec)scanf("%ld",&i)
#define fi first
#define se second

const int MOD = 1e9+7;
const ll INF = 1e18;
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, -1, 1};

int main(){
	ll n, d, a;
	scanf("%ld %ld %ld", &n, &d, &a);
	std::vector<P> xh(n);
	rep(i, 0, n) scanf("%ld %ld", &xh[i].fi, &xh[i].se);
	std::sort(be(xh));

	V b(n, 0);
	ll bius = 0, ans = 0;
	rep(i, 0, n){
		bius -= b[i];
		xh[i].se = std::max(0l, xh[i].se - bius * a);
		ll x = (xh[i].se + a - 1) / a;
		ans += x;
		bius += x;
		ll idx = std::upper_bound(be(xh), P(xh[i].fi+2*d,1e10)) - xh.begin();
		if(idx<n) b[idx] += x;
	}
	printf("%ld\n", ans);
	return 0;
}