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

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n,d,a,x,h,bomb=0,tot=0,ans=0;cin>>n>>d>>a;d*=2;
  vector<P> v(n);
  rep(i,n) { cin>>x>>h; v[i]=P{x,(h+a-1)/a}; }
  sort(v.begin(),v.end());
  queue<P> q;
  rep(i,n) {
    while(q.size() && q.front().first<v[i].first) {
      tot-=q.front().second;
      q.pop();
    }
    v[i].second-=tot;
    if(v[i].second>0) {
      bomb=v[i].second;
      ans+=bomb;
      tot+=bomb;
      q.emplace(v[i].first+d,bomb);
    }
  }
  print(ans);
  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;
#define ll long long
#define lvector vector<ll> 
typedef pair<ll,ll> P;
#define rep(i,n) for(ll (i)=0; i<(n); ++(i))
#define print(a) cout << (a) << endl

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n,d,a,x,h,bomb=0,tot=0,ans=0;cin>>n>>d>>a;d*=2;
  vector<P> v(n);
  rep(i,n) { cin>>x>>h; v[i]=P{x,(h+a-1)/a}; }
  sort(v.begin(),v.end());
  queue<P> q;
  rep(i,n) {
    while(q.size() && q.front().first<v[i].first) {
      tot-=q.front().second;
      q.pop();
    }
    v[i].second-=tot;
    if(v[i].second>0) {
      bomb=v[i].second;
      ans+=bomb;
      tot+=bomb;
      q.emplace(v[i].first+d,bomb);
    }
  }
  print(ans);
  return 0;
}
```