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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MOD1 1000000007
#define MOD2 998244353
#define LIMIT1 200002
#define LIMIT2 500002
typedef long long ll;
typedef long double ld;
typedef const void cv;
#define rep(i,n) for((i)=0;(i)<(n);(i)++)
#define per(i,n) for((i)=(n)-1;(i)>=0;(i)--)
#define max(a,b) ((a)>(b) ? (a) : (b))
#define min(a,b) ((a)<(b) ? (a) : (b))
#define zt(a,b) (max((a),(b))-min((a),(b)))
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll nPr(int n,int r){ll i,result=1;rep(i,r){result*=(ll)(n-i);}return result;}
ll nCr(int n,int r){ll i,result=1;if(n<r) return 0;rep(i,min(r,n-r)){result*=(ll)(n-i);result/=(ll)(i+1);}return result;}
#define nHr(n,r) nCr((int)((n)+(r)+1),(int)(r))
#define sankaku(x) (((x)*((x)+1))/2)
int dx[8]={1,0,-1,0,1,-1,-1,1};
int dy[8]={0,1,0,-1,1,1,-1,-1};
int upint(const void *a, const void *b) { return *(int *)a < *(int *)b ? -1 : *(int *)a > *(int *)b ? 1 : 0; }
int downint(const void *a, const void *b) { return *(int *)a < *(int *)b ? 1 : *(int *)a > *(int *)b ? -1 : 0; }
int upchar(const void* left, const void* right) {return strcmp((char *)left,(char *)right);}
int downchar(const void* left, const void* right) {return strcmp((char *)right,(char *)left);}
ll modpow(ll a, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1){
            res *= a; if(mod>=1) res %= mod;
        }
        a *= a; if(mod>=1) a %= mod;
        n >>= 1;
    }
    return res;
}
void initialize(){
    
}
ll x[LIMIT2]={0},h[LIMIT1]={0};
int c[LIMIT1]={0};
int tmp(cv *p,cv *q){
  return x[*(int *)p]-x[*(int *)q];
}
int main(void){
  initialize();
  ll n,m,i,j,k,result=0,d,a,l,r;
  char s[LIMIT1];
  ll sum[LIMIT1]={0};//累積和(攻撃回数)
  ll total=0;
  
  scanf("%lld %lld %lld",&n,&d,&a);
  rep(i,n){
    scanf("%lld %lld",&x[i],&h[i]);
    h[i] = (ll)ceill((ld)h[i]/a);//hp->攻撃数
    c[i]=(int)i;
  }
  qsort(c,n,sizeof(int),tmp);
  l=0;
  r=0;
  rep(l,n){
    total+=sum[l];
    //printf("%lld\n",(ll)h[c[l]]-total);
    if(h[c[l]]<=total){
      continue;//already fainted
    }
    while(r<n && x[c[r]]<=x[c[l]]+2*d){
      r++;
    }
    //printf("%lld %lld\n",l,r);
    sum[l]+=h[c[l]]-total;
    if(r<n){
      sum[r]-=h[c[l]]-total;
    }
    result+=h[c[l]]-total;
    total+=h[c[l]]-total;
    //printf("%d %d\n",(x[c[l]])+(int)d,h[c[l]]-total);
    if(l==r) r++;
    else{
    }
  }
  printf("%lld\n",result);
  return 0;
}