#include<bits/stdc++.h>
using namespace std;
const long long maxn=200010LL;
long long n,d,a,xx,yy,res=0LL,pre[maxn];
pair<long long,long long> v[maxn];
long long num(long long x)
{
	long long ans=x/a;
	if(x%a) ++ans;
	return ans;
}
int main()
{
	cin>>n>>d>>a;
	for(long long i=1LL;i<=n;i++) cin>>v[i].first>>v[i].second;
	sort(v+1LL,v+1LL+n);
	v[n+1]=make_pair(1e18,1e18);
	for(long long i=1LL;i<=n;i++)
	{
		pre[i]+=pre[i-1];
		long long di=v[i].second-pre[i];
		if(di<=0LL) continue;
		long long tmp=v[i].first+2*d; 
		long long ind=upper_bound(v+1LL,v+2LL+n,make_pair(tmp+1,0LL))-v-1;
		long long tim=num(di);
		res+=tim;
		long long att=tim*a;
		pre[i]+=att;
		pre[ind+1]-=att;
	}
	cout<<res<<endl;
	return 0;
}
### Prompt

Your challenge is to write 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;
const long long maxn=200010LL;
long long n,d,a,xx,yy,res=0LL,pre[maxn];
pair<long long,long long> v[maxn];
long long num(long long x)
{
	long long ans=x/a;
	if(x%a) ++ans;
	return ans;
}
int main()
{
	cin>>n>>d>>a;
	for(long long i=1LL;i<=n;i++) cin>>v[i].first>>v[i].second;
	sort(v+1LL,v+1LL+n);
	v[n+1]=make_pair(1e18,1e18);
	for(long long i=1LL;i<=n;i++)
	{
		pre[i]+=pre[i-1];
		long long di=v[i].second-pre[i];
		if(di<=0LL) continue;
		long long tmp=v[i].first+2*d; 
		long long ind=upper_bound(v+1LL,v+2LL+n,make_pair(tmp+1,0LL))-v-1;
		long long tim=num(di);
		res+=tim;
		long long att=tim*a;
		pre[i]+=att;
		pre[ind+1]-=att;
	}
	cout<<res<<endl;
	return 0;
}
```