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: Wrong Answer)

// https://atcoder.jp/contests/abc153/tasks/abc153_f
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define REP(i, n) FOR(i, 0, n)
#define REPR(i, n) for (int i = n - 1; i >= 0; i--)
#define FOR(i, s, n) for (int i = (s), i##_len = (n); i < i##_len; ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
#define DIV(a, b) ((a - 1) / b + 1)
const int INF = 1e10;
using pii = pair<int, int>;

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N, D, A;
    cin >> N >> D >> A;
    priority_queue<pii, vector<pii>, greater<pii>> q;
    REP(i, N) {
        int a, b;
        cin >> a >> b;
        q.push({a, b});
    }
    int ans = 0;
    while (!q.empty()) {
        auto t = q.top();
        q.pop();
        int end = t.first + 2 * D;
        priority_queue<pair<int, int>> r;
        while (end <= t.first) {
            if (t.second - D > 0) {
                r.push({t.first, t.second - D});
            }
            t = q.top();
            q.pop();
        }
        while (!r.empty()) {
            t = r.top();
            r.pop();
            q.push(t);
        }

        ans++;
    }
    cout << ans << endl;

    return 0;
}