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>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
#define rep(i,n) for (ll i = 0; i < (n); ++i)
#define vcout(v) rep(i,v.size()) cout << v[i] << " "; cout << endl
#define mcout(m) rep(j,m.size()){vcout(m[j]);}
const ll MOD = 1e9 + 7;

void mainsolve(){

	ll n,d,a;
	cin >> n >> d >> a;
	vector<P> xx(n);
	rep(i,n) cin >> xx[i].first >> xx[i].second;
	sort(xx.begin(), xx.end());

	queue<P> q; // 効くぎりぎりの座標と、その時点でのtotのpair
	ll tot = 0;
	ll ans = 0;
	rep(i,n){
		ll x = xx[i].first;
		ll h = xx[i].second;
		if(tot > 1e18) rep(i,tot) cout << i;
		h -= tot;
		// totで引きすぎた分を調整
		while(!q.empty() && q.front().first < x){
			h += q.front().second;
			q.pop();
		}
		
		// i番目の敵を爆風の左端にして倒す
		if(h > 0){
			ll cnt = 0;
			if(h%a==0) cnt = h/a;
			else cnt = h/a + 1;
			ans += cnt;
			tot += a * cnt;
			q.push(P(x+2*d,tot));
		}
		
		
	}

	cout << ans << endl;
	return;
}

int main(){
	
	//ll TC;cin>>TC;rep(I,TC){cout<<"case:"<<I+1<<"---"<<endl; // comment out!
		mainsolve();
	//} // comment out!
	return 0;
}