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>
#define mod 1000000007
#define MAXN 200000 + 5
#define ll long long
#define int long long
using namespace std;
 
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define all(arr) arr.begin() , arr.end()
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define rep(i , a , n) for(int i = a ; i < n ; i++)
#define s(arr) sort(all(arr))
#define r(arr) reverse(all(arr))
#define rs(arr) s(arr) ; r(arr)
#define con continue
#define endl "\n"
 
//debug
#define trace1(x)                cout<<#x<<": "<<x<<endl
#define trace2(x, y)             cout<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)          cout<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl

signed main(){
    FAST;
    int n , d , a;
    cin >> n >> d >> a;
    vector< pair <int , int> > arr(n);
    rep(i , 0 , n){
        cin >> arr[i].ff >> arr[i].ss;
    }
    sort(all(arr));
    int ans = 0;
    int i = 0 , j = 0;
    while(i < n){
        if (arr[i].ss <= 0){
            i++;
            continue;
        }
        j = max(j , i);
        while(j < n && ((arr[j].ff - arr[i].ff) <= (2 * d + 1))){
            j++;
        }
        //trace2(i , j);
        int req = arr[i].ss / a;
        ans += req;
        int fl = 0;
        i++;
        int j1 = i;
        while(j1 < j){
            arr[j1].ss -= req * a;
            j1++;
        }
        while(i < n && arr[i].ss <= 0)
            i++;
    }
    cout << ans << endl;
    return 0;
}