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)

#include<bits/stdc++.h>

int main(){
    using namespace std;
    unsigned long N, D, A;
    cin >> N >> D >> A;
    vector<pair<unsigned long, unsigned long>> enemies(N);
    for(auto& i : enemies){
        cin >> i.first >> i.second;
        (i.second += A - 1) /= A;
    }
    sort(enemies.begin(), enemies.end(), greater<>());
    unsigned long ans{0}, now{0};
    queue<pair<unsigned long, unsigned long>> pq;
    pq.emplace(0, 0);
    for(const auto& i : enemies){
        while(pq.front().first > i.first && (now += pq.front().second))pq.pop();
        auto k = i.second - min(i.second, ans - now);
        if(k)pq.emplace(i.first - min(i.first, D * 2), k);
        ans += k;
    }
    cout << ans << endl;
    return 0;
}