#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
  ll n,d,a;cin >> n >> d >> a;
  vector<pair<ll,ll>> p(n);
  for(int i=0;i<n;i++){
    ll x,h;cin >> x >> h;
    p[i]={x,(h+a-1)/a};
  }
  sort(p.begin(),p.end());
  vector<ll> x(n),y(n),c(n+1);
  for(int i=0;i<n;i++){
    x[i]=p[i].first;
    y[i]=p[i].second;
  }
  ll sum=0,ans=0;
  for(int i=0;i<n;i++){
    int itr=upper_bound(x.begin(),x.end(),x[i]+2*d)-x.begin();
    sum+=c[i];
    ll r=max(0LL,y[i]-sum);
    ans+=r;
    c[i+1]+=r;
    c[itr]-=r;
  }
  cout << ans << endl;
}
### 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>
using namespace std;
typedef long long ll;
int main(){
  ll n,d,a;cin >> n >> d >> a;
  vector<pair<ll,ll>> p(n);
  for(int i=0;i<n;i++){
    ll x,h;cin >> x >> h;
    p[i]={x,(h+a-1)/a};
  }
  sort(p.begin(),p.end());
  vector<ll> x(n),y(n),c(n+1);
  for(int i=0;i<n;i++){
    x[i]=p[i].first;
    y[i]=p[i].second;
  }
  ll sum=0,ans=0;
  for(int i=0;i<n;i++){
    int itr=upper_bound(x.begin(),x.end(),x[i]+2*d)-x.begin();
    sum+=c[i];
    ll r=max(0LL,y[i]-sum);
    ans+=r;
    c[i+1]+=r;
    c[itr]-=r;
  }
  cout << ans << endl;
}
```