#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
ll n,d,a,ans,sum,ab;
pii p[200005];
queue<pii> q;
int main(){
	n=read(),d=read()*2,a=read();
	for(int i=1;i<=n;i++)
		p[i].first=read(),p[i].second=read();
	sort(p+1,p+n+1);
	for(int i=1;i<=n;i++){
		while(!q.empty() and p[i].first>q.front().first)
			sum-=q.front().second,q.pop();
		ab=p[i].second-sum;
		if(ab<=0)continue;
		ab=(ab+a-1)/a;
		ans+=ab,sum+=ab*a;
		q.push(pii{p[i].first+d,ab*a});
	}
	cout << ans;
	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>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
ll n,d,a,ans,sum,ab;
pii p[200005];
queue<pii> q;
int main(){
	n=read(),d=read()*2,a=read();
	for(int i=1;i<=n;i++)
		p[i].first=read(),p[i].second=read();
	sort(p+1,p+n+1);
	for(int i=1;i<=n;i++){
		while(!q.empty() and p[i].first>q.front().first)
			sum-=q.front().second,q.pop();
		ab=p[i].second-sum;
		if(ab<=0)continue;
		ab=(ab+a-1)/a;
		ans+=ab,sum+=ab*a;
		q.push(pii{p[i].first+d,ab*a});
	}
	cout << ans;
	return 0;
}
```