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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define rep2(i, a, n) for(ll i = a; i < (ll)(n); i++)
#define memi cout << endl
#define kono(n) cout << fixed << setprecision(n)
#define all(c) (c).begin(), (c).end()
#define pb push_back
#define hina cout << ' '
#define in(n) cin >> n
#define in2(n, m) cin >> n >> m
#define in3(n, m, l) cin >> n >> m >> l
#define out(n) cout << n
const ll mei = (ll)1e9 + 7;

int main(){
  ll n, d, a, k, s, t;
  in3(n, d, a);
  vector<pair<ll, ll>> c(n);
  rep(i, n)
    in2(c[i].first, c[i].second);
  c.pb(make_pair(9999999999999, 999999999999999));
  sort(all(c));
  d = d * 2;
  vector<ll> f(n), e(n, 0);
  s = 0;
  t = 1;
  while(s < n){
    while(c[t].first <= c[s].first + d)
      t++;
    f[s] = t;
    s++;
  }
  s = 0; // damage
  k = 0;
  rep(i, n){
    s -= e[i];
    if(s >= c[i].first)
      continue;
    t = (c[i].second - s + a - 1) / a;
    s += t * a;
    e[f[i]] += t * a;
    k += t;
  }
  out(k);
  memi;
}