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>
using namespace std;
typedef unsigned long long ull;
typedef long long int ll;
#define ll long long int
#define dd double
#define MOD 1000000007
#define nl cout<<endl
#define rt ll Q[100005]={0}
#define mp make_pair
#define test() ull t;cin>>t;while(t--)
#define pb push_back
#define ff first    
#define ss second
#define all(v) v.begin(),v.end()
#define s(ar,n) sort(ar,ar+n)
#define rs(ar,n) sort(ar, ar+n, greater<ll>()) 
#define oa(a,n) for(ll i=0;i<n;i++)cout<<a[i]<<" ";nl
#define cn(a,n) for(ll i=0;i<n;i++)cin>>a[i];
#define ov(a,n) for(ll i=0;i<n;i++)cout<<a[i]<<" ";nl
#define ovp(a,n) for(ll i=0;i<n;i++)cout<<a[i].ff<<" "<<a[i].ss<<endl;
#define maxa(ar,N) *max_element(ar,ar+N)
#define mina(ar,N) *min_element(ar,ar+N)
#define fastio() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
typedef pair<int,int> pii;
ll powe(ll n,ll m){if(m==0)return 1;ll t=powe(n,m/2);if(m%2==0)return (t*t);return (((t*t))*n);}
ll mpowe(ll n,ll m){if(m==0)return 1;ll t=powe(n,m/2);t%=MOD;if(m%2==0)return (t*t)%MOD;return (((t*t)%MOD)*n)%MOD;}
ll logtwo(ll n){if(n==1)return 0;return logtwo(n/2)+1;}
ll A[200000+1];

int main()
{
    fastio();
    ll n,d,a,i,b,c,j,ne=0,de=0,ans=0,l,r,mid;
    cin >> n >> d >> a;
  // cout << n << d << a;
    pair<ll,ll>v[n];
    
        for (i=1; i<=n; i++) 
            {   cin >> v[i].first >> v[i].second;
            }
        sort(v+1,v+1+n);
        //cout << v[3].first;
        //cout << v[0].second;
        // de=0;
        for(i=1;i<=n;i++)
        {
            v[i].second-=de;
            //cout << v[i].second;
            if(v[i].second>0)
            {
                ne=(v[i].second+a-1)/a;
                de=de+ne*a;
                //cout << ne ;
                ans=ans+ne;
                l=i,r=n+1;
                //cout << l << r;
                while(l<r-1)
                {
                   mid=(l+r)/2;
                   //cout << mid;
                   if(v[mid].first<=2*d+v[i].first)
                        l=mid;
                   else
                        r=mid;
                }
               // cout << l;
                A[l]=A[l]+de;
            }
            de=de-A[i];
            //cout << A[2]<<endl;;
            //cout << v[1].second<<endl;
        }
        cout << ans;
}