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)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=int(a);i<int(b);i++)
typedef long long ll;



int main(){
    ll n,d,a;cin>>n>>d>>a;
    vector<pair<ll,ll> > v;
    ll cn=0;
    vector<pair<ll,ll> > e(n);
    rep(i,0,n)cin>>e[i].first>>e[i].second;
    sort(e.begin(),e.end());
    ll dm=0,ite=0;
    rep(i,0,n){
        if(v.size()!=0&&ite<v.size()){
            while(ite<v.size()&&v[ite].first<e[i].first){
                dm-=v[ite].second;
                ite++;
            }
        }
        ll nowh=e[i].second-dm;
        if(nowh){
            dm+=(nowh+a-1)/a*a;
            v.push_back(make_pair(e[i].first+2*d,(nowh+a-1)/a*a));
        }
        cn+=(nowh+a-1)/a;
        
    }
    cout<<cn<<endl;
}