#include<bits/stdc++.h>
using namespace std;
pair<long long,long long>xh[200009];
int n;
long long d,a,ans[200009];
int main() {
	scanf("%d%lld%lld",&n,&d,&a);
	for(int i=0;i<n;i++) scanf("%lld%lld",&xh[i].first,&xh[i].second),xh[i].second=(xh[i].second+a-1)/a;
	sort(xh,xh+n);
	for(int i=0;i<n;i++) {
		int pos=upper_bound(xh,xh+n,make_pair(xh[i].first-2*d-1,(long long)1e18))-1-xh;
		ans[i]=(i>0?ans[i-1]:0)+max(xh[i].second-(i>0?ans[i-1]:0)+(pos>=0?ans[pos]:0),(long long)0);
	}
	printf("%lld\n",ans[n-1]);
	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;
pair<long long,long long>xh[200009];
int n;
long long d,a,ans[200009];
int main() {
	scanf("%d%lld%lld",&n,&d,&a);
	for(int i=0;i<n;i++) scanf("%lld%lld",&xh[i].first,&xh[i].second),xh[i].second=(xh[i].second+a-1)/a;
	sort(xh,xh+n);
	for(int i=0;i<n;i++) {
		int pos=upper_bound(xh,xh+n,make_pair(xh[i].first-2*d-1,(long long)1e18))-1-xh;
		ans[i]=(i>0?ans[i-1]:0)+max(xh[i].second-(i>0?ans[i-1]:0)+(pos>=0?ans[pos]:0),(long long)0);
	}
	printf("%lld\n",ans[n-1]);
	return 0;
}
```