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)

/*
?!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!   Author: Zaki Uddin        !
!   Created : 11:41 30/08/20  !
?!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
*/

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
const ll mod= 1e9+7;
const double pi = 3.141592653589793238462643383279;

#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define popc __builtin_popcount
#define popcll __builtin_popcountll
#define Sort(c) sort((c).begin(),(c).end())
#define all(X) (X).begin(),(X).end()
#define Unique(X) (X).erase(unique(all(X)),(X).end())

#define elif else if

void solve()
{
    ll n,d,a;
    cin>>n>>d>>a;
    vector<pair<ll,ll>> v;
    for(ll i=0;i<n;i++){
        ll x,h;
        cin>>x>>h;
        v.push_back({x,ceil(h/a)});
    }

    Sort(v);

    ll pre[n+1]={0},extraBombs=0;

		ll ans=0;
		for(ll i=n-1;~i;i--)
		{
			pair<ll,ll> p={v[i].first-2*d,-mod};
			ll  lidx=lower_bound(all(v),p)-v.begin();
			ll  left=v[i].second+extraBombs;
            // cout<<lidx<<" ";
            // cout<<p.first<<" "<<p.second<<" "<<left<<" "<<extraBombs<<"\n";
			if(left)
			{
				ans+=left;
				pre[lidx]+=left;
				extraBombs-=left;
			}
			extraBombs+=pre[i];
		}
		cout<<ans;
}
signed main()
{
   fast;
   ll testCases=1;
//    cin>>testCases;
   while(testCases--)
   {
       solve();
       cout<<"\n";
   }
}