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 9.2.1) (judge verdict: Wrong Answer)

//g++ -std=gnu++14 a.cpp
//#include <bits/stdc++.h>

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <tuple>
#include <iomanip>
#include <random>
#include <math.h>

using namespace std;


#define ll long long
#define rep(i, n) for(ll i = 0; i < (n); i++)
ll MOD = 1e9 + 7;
int INF = 1 << 30;
ll INFL = 1LL << 60;





int main(){
   ll N,D,A;
   cin >> N >> D >> A;
   vector<pair<ll,ll>> X(200001);
   map<ll,ll> m;
   rep(i,N){
     ll a,b;
     cin >> a >> b;
     m[a] = b;
   }
   ll i = 0;

   for(auto x : m){
     X[i] = make_pair(x.first,x.second);
     i++;
   }
   ll ans = 0;

   ll start = X[0].first;
   ll hazi = start + 2*D , max = 0;
/*
   rep(i,N){
     cout << X[i].first << " ";
   }
   cout << endl;
   rep(i,N){
     cout << X[i].second << " ";
   }
   cout << endl;
*/
   ll j = 0;

   while(j < N){
     //cout << ans << endl;
     if(X[j].second > max && X[j].first <= hazi){
       ll add = (X[j].second- max -1+A)/A;
       ans += add;
       max = add*A;
       hazi = X[j].first + 2*D;
       j++;
     }else if(X[j].second <= max && X[j].first <= hazi){
       j++;
       continue;
     //}else if(X[j].second > max && X[j].first > hazi){
     }else{
       max = 0;
       ll add = (X[j].second - max -1+A)/A;
       ans += add;
       max = add*A;
       hazi = X[j].first + 2*D;
       j++;
     }//else{
    //   j++;
    //   hazi = X[j].first + 2*D;
    // }
   }

   cout << ans << endl;

}