#include<bits/stdc++.h>
using namespace std;
long long pre[200007];
int main(){
	ios::sync_with_stdio(false);//关同步 
	long long n,d,a;
	cin>>n>>d>>a;
	vector<pair<long long,long long> >v;
	for(int i=1;i<=n;++i){
		long long x,y;
		cin>>x>>y;
		long long z=(y-1)/a+1;
        v.push_back({x,z});
    }
    sort(v.begin(),v.end());
    long long ans=0;
    long long sum=0;
    int r=1;//打不到的最近范围 
    for(int i=0;i<n;++i){
		while(r<n&&v[r].first<=v[i].first+2*d)
			++r;
		sum-=pre[i];//sum为当前位置已经预先承受的伤害总量
		v[i].second-=sum;
		if(v[i].second<=0)
			continue;
		sum+=v[i].second;//累计伤害 
		pre[r]+=v[i].second;//r位置以前承受的伤害总量 
		ans+=v[i].second;
	}
	cout<<ans;
	return 0;
}
### Prompt

Please formulate 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;
long long pre[200007];
int main(){
	ios::sync_with_stdio(false);//关同步 
	long long n,d,a;
	cin>>n>>d>>a;
	vector<pair<long long,long long> >v;
	for(int i=1;i<=n;++i){
		long long x,y;
		cin>>x>>y;
		long long z=(y-1)/a+1;
        v.push_back({x,z});
    }
    sort(v.begin(),v.end());
    long long ans=0;
    long long sum=0;
    int r=1;//打不到的最近范围 
    for(int i=0;i<n;++i){
		while(r<n&&v[r].first<=v[i].first+2*d)
			++r;
		sum-=pre[i];//sum为当前位置已经预先承受的伤害总量
		v[i].second-=sum;
		if(v[i].second<=0)
			continue;
		sum+=v[i].second;//累计伤害 
		pre[r]+=v[i].second;//r位置以前承受的伤害总量 
		ans+=v[i].second;
	}
	cout<<ans;
	return 0;
}
```