#include<bits/stdc++.h>
#define int long long
using namespace std;
struct abcd{int x,h;}b[200001];
bool cmp(abcd i,abcd j){return i.x<j.x;}
int n,d,a,qn[200001],qx[200001],q1,q2,qs,ans;
signed main()
{
 scanf("%lld%lld%lld",&n,&d,&a);
 for(int i=1;i<=n;i++)
 {
  scanf("%lld%lld",&b[i].x,&b[i].h);
  b[i].h=(b[i].h-1)/a+1;
 }
 sort(b+1,b+n+1,cmp);
 for(int i=1;i<=n;i++)
 {
  while(qx[q1]<b[i].x&&q1<q2)qs-=qn[q1++];
  if(b[i].h>qs)
  {
   ans+=(qn[q2]=b[i].h-qs);
   qx[q2++]=b[i].x+2*d;
   qs=b[i].h;
  }
 }
 printf("%lld",ans);
 return 0;
}
### Prompt

Generate 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>
#define int long long
using namespace std;
struct abcd{int x,h;}b[200001];
bool cmp(abcd i,abcd j){return i.x<j.x;}
int n,d,a,qn[200001],qx[200001],q1,q2,qs,ans;
signed main()
{
 scanf("%lld%lld%lld",&n,&d,&a);
 for(int i=1;i<=n;i++)
 {
  scanf("%lld%lld",&b[i].x,&b[i].h);
  b[i].h=(b[i].h-1)/a+1;
 }
 sort(b+1,b+n+1,cmp);
 for(int i=1;i<=n;i++)
 {
  while(qx[q1]<b[i].x&&q1<q2)qs-=qn[q1++];
  if(b[i].h>qs)
  {
   ans+=(qn[q2]=b[i].h-qs);
   qx[q2++]=b[i].x+2*d;
   qs=b[i].h;
  }
 }
 printf("%lld",ans);
 return 0;
}
```