#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)n;i++)
typedef long long ll;
int main(void){
   int n;
   ll d,a;cin>>n>>d>>a;
   vector<pair<ll,ll>> p(n);
   rep(i,n)cin>>p[i].first>>p[i].second;
   sort(p.begin(),p.end());
   d*=2;
   ll tot=0;
   queue<pair<ll,ll>> q;
   ll ans=0;
   rep(i,n){
       ll x=p[i].first;
       ll h=p[i].second;
       while(q.size()&&q.front().first<x){
           tot-=q.front().second;
           q.pop();
       }
       h-=tot;
       if(h>0){
           ll num=(h+a-1)/a;
           ans+=num;
           ll damage=num*a;
           tot+=damage;
           q.emplace(x+d, damage);
       }
   }
   cout<<ans<<endl;
}
### Prompt

Please create a solution in CPP 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;
#define rep(i,n) for(int i=0;i<(int)n;i++)
typedef long long ll;
int main(void){
   int n;
   ll d,a;cin>>n>>d>>a;
   vector<pair<ll,ll>> p(n);
   rep(i,n)cin>>p[i].first>>p[i].second;
   sort(p.begin(),p.end());
   d*=2;
   ll tot=0;
   queue<pair<ll,ll>> q;
   ll ans=0;
   rep(i,n){
       ll x=p[i].first;
       ll h=p[i].second;
       while(q.size()&&q.front().first<x){
           tot-=q.front().second;
           q.pop();
       }
       h-=tot;
       if(h>0){
           ll num=(h+a-1)/a;
           ans+=num;
           ll damage=num*a;
           tot+=damage;
           q.emplace(x+d, damage);
       }
   }
   cout<<ans<<endl;
}
```