#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)

typedef long long ll;
typedef pair<int,int> P;

int main(){
	ll N,D,A;
	cin >> N >> D >> A;
	vector<P> E(N);
	rep(i,N){ int x,h; cin >> x >> h; E[i]=P(x,h); }
	sort(E.begin(),E.end());

	ll cnt = 0;
	int num = 0;
	ll range = D*2;
	ll dam = 0;
	queue<pair<ll,ll>> q;
	while(num < N){
		int loc = E[num].first;
		int hp = E[num].second;
		
		while(q.size() &&  q.front().first < loc){
			dam -= q.front().second;
			q.pop();
		}
		hp -= dam;
		if(hp>0){
			ll c = (hp+A-1)/A;
			cnt += c;
			ll damage = c * A;
			dam += damage;
			q.emplace(loc+range, damage);
		}
		num++;
	}

	cout << cnt << endl;
	return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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.

Examples

Input

3 3 2
1 2
5 4
9 2


Output

2


Input

9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5


Output

5


Input

3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000


Output

3000000000

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)

typedef long long ll;
typedef pair<int,int> P;

int main(){
	ll N,D,A;
	cin >> N >> D >> A;
	vector<P> E(N);
	rep(i,N){ int x,h; cin >> x >> h; E[i]=P(x,h); }
	sort(E.begin(),E.end());

	ll cnt = 0;
	int num = 0;
	ll range = D*2;
	ll dam = 0;
	queue<pair<ll,ll>> q;
	while(num < N){
		int loc = E[num].first;
		int hp = E[num].second;
		
		while(q.size() &&  q.front().first < loc){
			dam -= q.front().second;
			q.pop();
		}
		hp -= dam;
		if(hp>0){
			ll c = (hp+A-1)/A;
			cnt += c;
			ll damage = c * A;
			dam += damage;
			q.emplace(loc+range, damage);
		}
		num++;
	}

	cout << cnt << endl;
	return 0;
}

```