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: Time Limit Exceeded)

#include <stdio.h>
#include <iostream>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>

#include <iomanip>

using namespace std;

#define rep(i,x) for(int i=0;i<x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define FIND(x, target) find(x.begin(),x.end(),target)
#define pb(x) push_back(x)
#define pob(x) pop_back(x)
#define output(x) cout<<x<<endl
#define ERASE(x, n) x.erase(x.begin()+n)
// #define int long long
// #define P pair<int, int>
typedef long long ll;

const long long int MOD = 1e9+7;
const ll INF = 1e17;
// ll dp[100005][13];
ll dp[10005];

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

signed main(void) {
  ll N, D, A;
  cin >> N >> D >> A;
  vector<pair<ll, ll> > pairs(N);
  rep(i, N) {
    ll X, H;
    cin >> X >> H;
    pairs[i] = make_pair(X, H);
  }

  SORT(pairs);

  ll ans = 0;
  for(int i=N-1; i>=0; i--) {
    while (pairs[i].second > 0) {
      ll times = (pairs[i].second-1)/A + 1;
      ans += times;
      ll pos = pairs[i].first;
      ll zero_pos = pos;
      bool flag = false;
      for (int j=i; j>=0; j--) {
        if (pairs[j].first == zero_pos) {
          pairs[j].second = 0;
          if (!flag) {
            i = j;
          }
        } else if (pairs[j].first >= pos - 2*D) {
          pairs[j].second -= A*times;
          if (pairs[j].second > 0) {
            flag = true;
          } else {
            zero_pos = pairs[j].first;
          }
          if (!flag) {
            i = j;
          }
        } else {
          break;
        }
      }
    }
  }

  cout << ans << endl;
}