#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
ll n,d,m;
ll c[200010];
pair<ll,ll> a[200010];
int main(){
	cin>>n>>d>>m;
	for(int i=1;i<=n;i++)cin>>a[i].first>>a[i].second;
	sort(a+1,a+n+1);
	ll temp=0;
	ll ans=0;
	d*=2;
	for(ll i=1;i<=n;i++){
		temp-=c[i];
		a[i].second-=temp;
		if(a[i].second>0){
			ll t=(a[i].second+m-1)/m;
			ans+=t;
			temp+=m*t;
			int l=0,r=n+1;
			while(l<r){
				int mid=(l+r)/2;
				if(a[mid].first>a[i].first+d)r=mid;
				else l=mid+1;
			}
			if(l!=n+1)c[l]+=m*t;
		}
	}
	cout<<ans<<endl;
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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>
typedef long long ll;
using namespace std;
ll n,d,m;
ll c[200010];
pair<ll,ll> a[200010];
int main(){
	cin>>n>>d>>m;
	for(int i=1;i<=n;i++)cin>>a[i].first>>a[i].second;
	sort(a+1,a+n+1);
	ll temp=0;
	ll ans=0;
	d*=2;
	for(ll i=1;i<=n;i++){
		temp-=c[i];
		a[i].second-=temp;
		if(a[i].second>0){
			ll t=(a[i].second+m-1)/m;
			ans+=t;
			temp+=m*t;
			int l=0,r=n+1;
			while(l<r){
				int mid=(l+r)/2;
				if(a[mid].first>a[i].first+d)r=mid;
				else l=mid+1;
			}
			if(l!=n+1)c[l]+=m*t;
		}
	}
	cout<<ans<<endl;
	return 0;
}
```