#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 1LL<<62
#define inf 1000000007

int main() {
	ll n,d,a;
	cin>>n>>d>>a;
	vector<pair<ll,ll>>ch;
	for(ll i=0;i<n;i++){
		ll x,y;
		cin>>x>>y;
		ch.push_back(make_pair(x,y));
	}
	sort(ch.begin(),ch.end());
	queue<pair<ll,ll>>q;
	ll ans=0;
	ll dm=0;
	for(ll i=0;i<n;i++){
		ll nowx=ch[i].first;
		while(q.size()!=0&&q.front().first<nowx){
			dm-=q.front().second;
			q.pop();
		}
		ch[i].second-=dm;
		if(ch[i].second>0){
			ans+=(ch[i].second+a-1)/a;
			ll dm1=((ch[i].second+a-1)/a)*a;
			dm+=dm1;
			q.push(make_pair(nowx+2*d,dm1));
		}
	}
	cout <<ans;
	// your code goes here
	return 0;
}
### Prompt

Your task is to create a CPP solution to 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;
typedef long long ll;
#define INF 1LL<<62
#define inf 1000000007

int main() {
	ll n,d,a;
	cin>>n>>d>>a;
	vector<pair<ll,ll>>ch;
	for(ll i=0;i<n;i++){
		ll x,y;
		cin>>x>>y;
		ch.push_back(make_pair(x,y));
	}
	sort(ch.begin(),ch.end());
	queue<pair<ll,ll>>q;
	ll ans=0;
	ll dm=0;
	for(ll i=0;i<n;i++){
		ll nowx=ch[i].first;
		while(q.size()!=0&&q.front().first<nowx){
			dm-=q.front().second;
			q.pop();
		}
		ch[i].second-=dm;
		if(ch[i].second>0){
			ans+=(ch[i].second+a-1)/a;
			ll dm1=((ch[i].second+a-1)/a)*a;
			dm+=dm1;
			q.push(make_pair(nowx+2*d,dm1));
		}
	}
	cout <<ans;
	// your code goes here
	return 0;
}
```