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: Accepted)

//#pragma GCC optimize("Ofast")
//#pragma GCC target ("avx2")
//#pragma GCC optimization ("O3")
//#pragma GCC optimization ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//typedef tree < int , null_type, less <int>, rb_tree_tag, tree_order_statistics_node_update> TREE;
#define ll long long int
#define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define pii pair<int,int>
#define pss pair<short,short>
#define pld pair<long double,long double >
#define ld long double
#define piii  pair<pii,int>
#define vii vector<pair<int,int> >
#define st first
#define nd second
#define pll pair<ll,ll>
#define speed ios::sync_with_stdio(false);cin.tie();cout.tie();
#define M_PI 3.14159265358979323846
//#define int long long
const int mod=1000000007;
//const int mod=1009;
//const int mod=998244353;
const int inf=1000000009;
const long long INF=1000000000000000009;
const long long big=1000000000000000;
const long double eps=0.000000000000000000001;
using namespace std;
pll T[2000005];
queue<pll > K;
int main()
{
    speed
    ll n,d,a;
    cin>>n>>d>>a;
    d*=2;
    for(int i=1;i<=n;i++)
        cin>>T[i].st>>T[i].nd;
    sort(T+1,T+n+1);
    ll suma=0,wynik=0;
    for(int i=1;i<=n;i++)
    {
        while(!K.empty()&&T[i].st-d>K.front().st)
        {
            suma-=K.front().nd;
            K.pop();
        }
        if(suma>=INF/a||suma*a>=T[i].nd)
            continue;
        ll ile=(T[i].nd-suma*a)/a;
        if((T[i].nd-suma*a)%a!=0)
            ile++;
        wynik+=ile;
        K.push(mp(T[i].st,ile));
        suma+=ile;
    }
    cout<<wynik;
    return 0;
}