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 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
#define int long long
using namespace std;

class Pair
{
    public:
    int x; ;
    int h;
    
};

bool comp(Pair &a, Pair &b)
{
    return a.x<b.x;
}


signed main()
{
	int n; cin>>n;
	int d; cin>>d;
	int a; cin>>a;
	
	Pair arr[n];
	for(int i=0;i<n;i++)
	{
	    cin>>arr[i].x>>arr[i].h;
	}
	sort(arr,arr+n,comp);
// 	for(int i=0;i<n;i++)
// 	{
// 	   cout<<arr[i].x<<" "<<arr[i].h<<endl;
// 	}
	
	stack<Pair> st;
	
	for(int i=0;i<n;i++)
	{
	   stack<Pair> temp; 
	   while(!st.empty()&&arr[i].h>0)
	   {   
	       
	       if(st.top().x-d<= arr[i].x &&st.top().x+d>=arr[i].x)
	       {
	           arr[i].h-=(st.top().h)*a;
	       }
	       temp.push(st.top());
	       st.pop();
	   }
	   
	   while(temp.empty()==false)
	   {
	       st.push(temp.top());
	       temp.pop();
	   }
	   
	   if(arr[i].h>0)
	   {
	      
	      int quo=arr[i].h/a;
	      int rem=arr[i].h%a;
	      int cnt=quo;
	      if(rem>0)  cnt=quo+1;
	      
	      Pair temp;
	      temp.x=arr[i].x+d;
	      temp.h=cnt;
	      
	      st.push(temp);
	      
	      
	   }
	   
	   
	   
	   
	    
	}
	
	int ans=0;
    while( st.empty()==false)
    {
       // cout<<st.top().x<<" "<<st.top().h<<endl;
        ans+=st.top().h;
        st.pop();
    }
    cout<<ans<<endl;
	
	
	
	
	
	return 0;
}