#include<bits/stdc++.h>
using namespace std;
using ll=long long;


int main(){
  ll n,d,a;
  cin >> n >> d >> a;
  vector<pair<ll,ll>> xh(n);
  for(ll i=0;i<n;++i){
    ll x,h;
    cin >> x >> h;
    xh[i]=make_pair(x,h);
  }

  sort(xh.begin(),xh.end());
  d=d*2;
  ll ans=0;
  ll tot=0;
  queue<pair<ll,ll>> q;
  for(ll i=0;i<n;++i){
    ll x=xh[i].first;
    ll h=xh[i].second;
    while(q.size()!=0 && q.front().first<x){
      tot-=q.front().second;
      q.pop();
    }
    h-=tot;
    if(h>0){
      ll num=(h+a-1)/a;
      ll damage=num*a;
      ans+=num;
      tot+=damage;
      q.push(make_pair(x+d,damage));
    }
  }

  cout << ans << endl;
  return 0;

}

### 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;
using ll=long long;


int main(){
  ll n,d,a;
  cin >> n >> d >> a;
  vector<pair<ll,ll>> xh(n);
  for(ll i=0;i<n;++i){
    ll x,h;
    cin >> x >> h;
    xh[i]=make_pair(x,h);
  }

  sort(xh.begin(),xh.end());
  d=d*2;
  ll ans=0;
  ll tot=0;
  queue<pair<ll,ll>> q;
  for(ll i=0;i<n;++i){
    ll x=xh[i].first;
    ll h=xh[i].second;
    while(q.size()!=0 && q.front().first<x){
      tot-=q.front().second;
      q.pop();
    }
    h-=tot;
    if(h>0){
      ll num=(h+a-1)/a;
      ll damage=num*a;
      ans+=num;
      tot+=damage;
      q.push(make_pair(x+d,damage));
    }
  }

  cout << ans << endl;
  return 0;

}

```