#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,d,a,h[200004],x[200004];
pair<ll,ll> peko[200004];
ll decr[200004],now;
ll ans=0;
int main(){
	cin>>n>>d>>a;
	for (ll i=1;i<=n;i++) cin>>peko[i].first>>peko[i].second;
	sort (peko+1,peko+1+n);
	for (ll i=1;i<=n;i++) x[i]=peko[i].first,h[i]=peko[i].second;
	for (ll i=1;i<=n;i++){
		if (now<h[i]){
			ll g=(h[i]-now+a-1)/a;
			ans+=g; 
			ll go=upper_bound(x+1,x+1+n,x[i]+2*d)-x;
			go--;
			decr[go]+=a*g;
			now+=a*g;
		}
		now-=decr[i];
	}
	cout<<ans;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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;
ll n,d,a,h[200004],x[200004];
pair<ll,ll> peko[200004];
ll decr[200004],now;
ll ans=0;
int main(){
	cin>>n>>d>>a;
	for (ll i=1;i<=n;i++) cin>>peko[i].first>>peko[i].second;
	sort (peko+1,peko+1+n);
	for (ll i=1;i<=n;i++) x[i]=peko[i].first,h[i]=peko[i].second;
	for (ll i=1;i<=n;i++){
		if (now<h[i]){
			ll g=(h[i]-now+a-1)/a;
			ans+=g; 
			ll go=upper_bound(x+1,x+1+n,x[i]+2*d)-x;
			go--;
			decr[go]+=a*g;
			now+=a*g;
		}
		now-=decr[i];
	}
	cout<<ans;
}
```