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)

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<ll,ll> P;
typedef vector<P> vp;
typedef vector<string> vs;
typedef vector<bool> vb;

const ll mod=1e9+7;
const ll inf=1e16;

#define rep(i,m,n) for(ll i=m;i<n;i++)
#define repr(i,m,n) for(ll i=m-1;i>=n;i--)
#define fi first
#define se second
#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define eb(x) emplace_back(x)
#define all(x) x.begin(),x.end()
#define gll greater<ll>()

const ll nmax=200005;
ll n;
vll bit(nmax,0);


void add(ll a, ll w){
  for(ll x=a;x<n;x|=x+1) bit[x]+=w;
}

ll sum(ll a){
 ll ret=0;
 for(ll x=a-1;x>=0;x=(x&(x+1))-1) ret+=bit[x];
 return ret;
}

void solve(){
  ll d,a;
  cin >> n >> d >> a;
  vp p(n);
  rep(i,0,n) cin >> p[i].fi >> p[i].se;
  sort(all(p));
  d*=2;
  
  queue<P> q;
  ll ans=0;
  rep(i,0,n){
   ll x=p[i].fi;
   ll h=p[i].se; 
   h-=sum(i); 
   if(h<=0) continue;
     ll num=(h+a-1)/a;
     ans+=num;
     add(i,num*a);
    
   ll rx=min(x+d,nmax-5); 
   ll j=upper_bound(all(p),P(rx+1,-1))-p.begin();
   add(j,-num*a);
  }
  cout << ans << endl;
}

int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);
  solve();
}