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



int main(){
    ll n,d,a;cin>>n>>d>>a;
    queue<pair<ll,ll> > q;
    ll cn=0;
    vector<pair<ll,ll> > e(n);
    rep(i,0,n)cin>>e[i].first>>e[i].second;
    sort(e.begin(),e.end());
    ll dm=0;
    rep(i,0,n){
        while(q.size()&&q.front().first<e[i].first){
            dm-=q.front().second;
            q.pop();
        }
        ll nowh=e[i].second-dm;
        if(nowh>0){
            dm+=(nowh+a-1)/a*a;
            q.push(make_pair(e[i].first+2*d,(nowh+a-1)/a*a));
            cn+=(nowh+a-1)/a;
        }
        
    }
    cout<<cn<<endl;
}
    








### Prompt

Please provide a cpp coded solution 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;
#define rep(i,a,b) for(int i=int(a);i<int(b);i++)
typedef long long ll;



int main(){
    ll n,d,a;cin>>n>>d>>a;
    queue<pair<ll,ll> > q;
    ll cn=0;
    vector<pair<ll,ll> > e(n);
    rep(i,0,n)cin>>e[i].first>>e[i].second;
    sort(e.begin(),e.end());
    ll dm=0;
    rep(i,0,n){
        while(q.size()&&q.front().first<e[i].first){
            dm-=q.front().second;
            q.pop();
        }
        ll nowh=e[i].second-dm;
        if(nowh>0){
            dm+=(nowh+a-1)/a*a;
            q.push(make_pair(e[i].first+2*d,(nowh+a-1)/a*a));
            cn+=(nowh+a-1)/a;
        }
        
    }
    cout<<cn<<endl;
}
    








```