#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair <int, int> P;
int main(){

  ll n,d,a;
  cin >>n>>d>>a;
  vector<pair<ll,ll> >x;
  for (ll i=0; i<n; i++){
    ll x1,h1;
    cin>>x1>>h1;
    x.push_back(make_pair(x1,h1));
  }
  queue <pair<ll, ll> > q;
  ll cnt=0;
  ll damage=0;
  sort(x.begin(), x.end());


  for (ll i=0; i<n; i++){

    while(q.size() && q.front().first<x[i].first){
      damage-=q.front().second;
      q.pop();
    }
    if (x[i].second<=damage) continue;


    ll attack=ceil(double(x[i].second-damage)/a);
    cnt+=attack;
    q.push(make_pair(x[i].first+2*d,attack*a));

    damage+=attack*a;


}
cout<<cnt<<endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
typedef pair <int, int> P;
int main(){

  ll n,d,a;
  cin >>n>>d>>a;
  vector<pair<ll,ll> >x;
  for (ll i=0; i<n; i++){
    ll x1,h1;
    cin>>x1>>h1;
    x.push_back(make_pair(x1,h1));
  }
  queue <pair<ll, ll> > q;
  ll cnt=0;
  ll damage=0;
  sort(x.begin(), x.end());


  for (ll i=0; i<n; i++){

    while(q.size() && q.front().first<x[i].first){
      damage-=q.front().second;
      q.pop();
    }
    if (x[i].second<=damage) continue;


    ll attack=ceil(double(x[i].second-damage)/a);
    cnt+=attack;
    q.push(make_pair(x[i].first+2*d,attack*a));

    damage+=attack*a;


}
cout<<cnt<<endl;
}

```