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")
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define double long double
using datas=pair<ll,ll>;
using ddatas=pair<double,double>;
using tdata=pair<ll,datas>;
using vec=vector<ll>;
using mat=vector<vec>;
using pvec=vector<datas>;
using pmat=vector<pvec>;
#define For(i,a,b) for(i=a;i<(ll)b;i++)
#define bFor(i,a,b) for(i=a;i>=(ll)b;i--)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)
#define brep(i,N) bFor(i,N-1,0)
#define all(v) (v).begin(),(v).end()
#define allr(v) (v).rbegin(),(v).rend()
#define vsort(v) sort(all(v))
#define vrsort(v) sort(allr(v))
#define endl "\n"
#define pb push_back
#define print(v) cout<<v<<endl
#define printyes cout<<"Yes"<<endl
#define printno cout<<"No"<<endl
#define printYES cout<<"YES"<<endl
#define printNO cout<<"NO"<<endl
#define output(v) do{bool f=0;for(auto outi:v){cout<<(f?" ":"")<<outi;f=1;}cout<<endl;}while(0)
const ll mod=1000000007;
const ll inf=1LL<<60;
const double PI = acos(-1);
const double eps = 1e-9;
template<class T> inline bool chmax(T& a,T b){bool x=a<b;if(x)a=b;return x;} 
template<class T> inline bool chmin(T& a,T b){bool x=a>b;if(x)a=b;return x;} 

void startupcpp(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(15);
}

double distance(ddatas x,ddatas y){
  double a=x.first-y.first,b=x.second-y.second;
  return sqrt(a*a+b*b);
}

ll modinv(ll a) {
  ll b=mod,u=1,v=0,t;
  while(b){
    t=a/b;
    a-=t*b; swap(a,b);
    u-=t*v; swap(u,v);
  }
  return (u+mod)%mod;
}

ll moddevide(ll a,ll b){return (a*modinv(b))%mod;}

vec modncrlistp,modncrlistm;

ll modncr(ll n,ll r){
  ll i,size=modncrlistp.size();
  if(size<=n){
    modncrlistp.resize(n+1);
    modncrlistm.resize(n+1);
    if(!size){
      modncrlistp[0]=modncrlistm[0]=1;
      size++;
    }
    For(i,size,n+1){
      modncrlistp[i]=modncrlistp[i-1]*i%mod;
      modncrlistm[i]=modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n]*modncrlistm[r]%mod*modncrlistm[n-r]%mod;
}

ll modpow(ll a,ll n){
  ll res=1;
  while(n){
    if(n&1)res=res*a%mod;
    a=a*a%mod;
    n>>=1;
  }
  return res;
}

ll gcd(ll a,ll b){if(!b)return a;return (a%b==0)?b:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

ll countdigits(ll n){
  ll ans=0;
  while(n){n/=10;ans++;}
  return ans;
}

ll sumdigits(ll n){
  ll ans=0;
  while(n){ans+=n%10;n/=10;}
  return ans;
}

struct LazySegmentTree{
  using func=function<ll(ll,ll)>;
  func f,g;
  ll N=1,E;
  vec node,lazy;
  vector<bool> vis;
  LazySegmentTree(vec& v,func F,func G,ll Ie):f(F),g(G),E(Ie){
    ll i=v.size();
    while(i){
      N<<=1;i>>=1;
    }
    node.resize(N*2-1,E);
    lazy.resize(N*2-1);
    vis.resize(N*2-1,false);
    rep(i,v.size()){
      node[N+i-1]=v[i];
    }
    brep(i,N-1){
      node[i]=f(node[i*2+1],node[i*2+2]);
    }
  }
  void eval(ll x){
    if(!vis[x])return;
    node[x]=g(node[x],lazy[x]);
    vis[x]=false;
    if(x<N-1){
      if(vis[x+x+1])lazy[x+x+1]=g(lazy[x],lazy[x+x+1]);
      else{
        lazy[x+x+1]=lazy[x];
        vis[x+x+1]=true;
      }
      if(vis[x+x+2])lazy[x+x+2]=g(lazy[x],lazy[x+x+2]);
      else{
        lazy[x+x+2]=lazy[x];
        vis[x+x+2]=true;
      }
    }
  }
  void update(ll a,ll b,ll x,ll now,ll l,ll r){
    eval(now);
    if((b<=l)|(r<=a))return;
    if((a<=l)&(r<=b)){
      vis[now]=true;lazy[now]=x;
      eval(now);
    }else{
      update(a,b,x,now+now+1,l,(l+r)>>1);
      update(a,b,x,now+now+2,(l+r)>>1,r);
      node[now]=f(node[now+now+1],node[now+now+2]);
    }
  }
  // +x in all[a,b);
  void add(ll a,ll b,ll x){
    update(a,b,x,0,0,N);
  }
  ll call(ll a,ll b,ll now,ll l,ll r){
    eval(now);
    if((b<=l)|(r<=a))return E;
    if((a<=l)&(r<=b))return node[now];
    return f(call(a,b,now+now+1,l,(l+r)>>1),call(a,b,now+now+2,(l+r)>>1,r));
  }
  //get min in[a,b);
  ll getmin(ll a,ll b){
    return call(a,b,0,0,N);
  }
};


int main(){
  // startupcpp();
  // ll test;cin>>test;while(test--){
  ll i,j,N,D,A,t,ans=0;
  cin>>N>>D>>A;
  D*=2;
  pvec v(N);
  vec h(N),x(N);
  rep(i,N)cin>>v[i].first>>v[i].second;
  vsort(v);
  rep(i,N){
    h[i]=v[i].second;
    x[i]=v[i].first;
  }
  LazySegmentTree tree(h,[](ll a,ll b){return max(a,b);},[](ll a,ll b){return a+b;},0);
  rep(i,N){
    j=tree.getmin(i,i+1);
    //print(j);
    if(j<=0)continue;
    auto itr=upper_bound(all(x),x[i]+D);
    t=(j+A-1)/A;
    ans+=t;
    tree.add(i,itr-x.begin(),-t*A);
  }
  print(ans);
  return 0;
}