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

int main()
{
ll n , d , a ;
cin >> n >> d >> a ;
vector<pair<ll,ll>> v(n);
rep(i,n){
  cin >> v[i].first >> v[i].second;
  v[i].second=(v[i].second+a-1)/a;
}
sort(v.begin(),v.end());
ll at=0;
queue<pair<ll, ll>> q;
rep(i,n){
  while(!q.empty()){
    auto r=q.front();
    if(r.first<v[i].first){
      q.pop();
      at-=r.second;
    }
    else break;
  }
  if(v[i].second>at){
    ans+=v[i].second-at;
    q.push({v[i].first+2*d,v[i].second-at});
    at+=v[i].second-at;
  }
}
cout << ans << endl;
return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n , d , a ;
cin >> n >> d >> a ;
vector<pair<ll,ll>> v(n);
rep(i,n){
  cin >> v[i].first >> v[i].second;
  v[i].second=(v[i].second+a-1)/a;
}
sort(v.begin(),v.end());
ll at=0;
queue<pair<ll, ll>> q;
rep(i,n){
  while(!q.empty()){
    auto r=q.front();
    if(r.first<v[i].first){
      q.pop();
      at-=r.second;
    }
    else break;
  }
  if(v[i].second>at){
    ans+=v[i].second-at;
    q.push({v[i].first+2*d,v[i].second-at});
    at+=v[i].second-at;
  }
}
cout << ans << endl;
return 0;
}
```