#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair <ll,ll> P;

int main(){
  ll n,d,a;
  cin>>n>>d>>a;
  P xh[200005];
  rep(i,n) cin>>xh[i].first>>xh[i].second;
  sort(xh,xh+n);
  ll x[200005],h[200005];
  rep(i,n){
    x[i]=xh[i].first;
    h[i]=xh[i].second;
  }
  vector<ll> sum(n+1,0);
  ll att_cnt=0;
  rep(i,n){
    if(i>0) sum[i]+=sum[i-1];
    h[i]-=sum[i];
    if(h[i]>0){
      ll att=(h[i]+a-1)/a*a;
      sum[i]+=att;
      att_cnt+=(h[i]+a-1)/a;
      sum[upper_bound(x,x+n,x[i]+2*d)-x]-=att;
    }
  }
  cout<<att_cnt<<endl;
  return 0;
}
### Prompt

Your task is to create 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 rep(i,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair <ll,ll> P;

int main(){
  ll n,d,a;
  cin>>n>>d>>a;
  P xh[200005];
  rep(i,n) cin>>xh[i].first>>xh[i].second;
  sort(xh,xh+n);
  ll x[200005],h[200005];
  rep(i,n){
    x[i]=xh[i].first;
    h[i]=xh[i].second;
  }
  vector<ll> sum(n+1,0);
  ll att_cnt=0;
  rep(i,n){
    if(i>0) sum[i]+=sum[i-1];
    h[i]-=sum[i];
    if(h[i]>0){
      ll att=(h[i]+a-1)/a*a;
      sum[i]+=att;
      att_cnt+=(h[i]+a-1)/a;
      sum[upper_bound(x,x+n,x[i]+2*d)-x]-=att;
    }
  }
  cout<<att_cnt<<endl;
  return 0;
}
```