#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;


int main(void){
  int n,d,i,x,h;
  ll a;
  P m[200005];
  
  cin >> n >> d >> a;
  for (i=1;i<=n;i++){
    cin >> x >> h;
    m[i]=P(x,h);
  }
  
  sort(m+1,m+n+1);
  
  int l,r,ko[200005]={};
  ll ru[200005];
  r=2;
  ru[0]=0;
  for (l=1;l<=n;l++){
    if (ko[l]==0) ko[l]=l;
    while (r<=n && m[r].first-m[l].first <= d*2){
      ko[r]=l;
      r++;
    }
    
    m[l].second-=a*(ru[l-1]-ru[ko[l]-1]);
    
    if (m[l].second<=0) ru[l]=0;
    else ru[l]=(m[l].second+a-1)/a;
    
    ru[l]+=ru[l-1];
  }
  
  cout << ru[n] << endl;
  
  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 <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;


int main(void){
  int n,d,i,x,h;
  ll a;
  P m[200005];
  
  cin >> n >> d >> a;
  for (i=1;i<=n;i++){
    cin >> x >> h;
    m[i]=P(x,h);
  }
  
  sort(m+1,m+n+1);
  
  int l,r,ko[200005]={};
  ll ru[200005];
  r=2;
  ru[0]=0;
  for (l=1;l<=n;l++){
    if (ko[l]==0) ko[l]=l;
    while (r<=n && m[r].first-m[l].first <= d*2){
      ko[r]=l;
      r++;
    }
    
    m[l].second-=a*(ru[l-1]-ru[ko[l]-1]);
    
    if (m[l].second<=0) ru[l]=0;
    else ru[l]=(m[l].second+a-1)/a;
    
    ru[l]+=ru[l-1];
  }
  
  cout << ru[n] << endl;
  
  return 0;
}
```