#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<ll,ll>P;
vector<P>vec;
ll sum[200009];

int main(){
  ll n,d,a;
  ll x[200009];
  ll ans=0;

  cin>>n>>d>>a;
  for(int i=0;i<n;i++){
    ll h;
    cin>>x[i]>>h;
    vec.push_back(P(x[i],h));
  }
  sort(vec.begin(),vec.end());
  sort(x,x+n);

  for(int i=0;i<n;i++){
    if(i>0)sum[i]+=sum[i-1];
    int nx=vec[i].first;
    int nh=max(vec[i].second-sum[i],0ll);
    int idx=upper_bound(x,x+n,nx+d*2)-x;
    int cnt=nh/a;
    if(nh%a!=0)cnt++;
    sum[i]+=a*cnt;
    sum[idx]-=a*cnt;
    ans+=cnt;
  }

  cout<<ans<<endl;
  return(0);
}

### Prompt

In CPP, your task is to solve 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 ll long long
typedef pair<ll,ll>P;
vector<P>vec;
ll sum[200009];

int main(){
  ll n,d,a;
  ll x[200009];
  ll ans=0;

  cin>>n>>d>>a;
  for(int i=0;i<n;i++){
    ll h;
    cin>>x[i]>>h;
    vec.push_back(P(x[i],h));
  }
  sort(vec.begin(),vec.end());
  sort(x,x+n);

  for(int i=0;i<n;i++){
    if(i>0)sum[i]+=sum[i-1];
    int nx=vec[i].first;
    int nh=max(vec[i].second-sum[i],0ll);
    int idx=upper_bound(x,x+n,nx+d*2)-x;
    int cnt=nh/a;
    if(nh%a!=0)cnt++;
    sum[i]+=a*cnt;
    sum[idx]-=a*cnt;
    ans+=cnt;
  }

  cout<<ans<<endl;
  return(0);
}

```