#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
pair<ll,ll> p[200005];
ll n,d,a,ans=0;
int main(){
    cin>>n>>d>>a;
    for(int i=1;i<=n;i++){
        cin>>p[i].first>>p[i].second;
        p[i].second=(p[i].second+a-1)/a;
    }
    sort(p+1,p+1+n);
    ll pos=1,mx=p[pos].second;ans+=mx;
    for(int i=2;i<=n;i++){
        while(p[i].first-p[pos].first>2*d){
            mx-=p[pos].second;
            pos++;
        }
        if(p[i].second<=mx) p[i].second=0;
        else p[i].second-=mx,mx+=p[i].second;
        ans+=p[i].second;
    }
    cout<<ans<<endl;
}                     
### Prompt

Please formulate a cpp solution 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;
typedef long long ll;
pair<ll,ll> p[200005];
ll n,d,a,ans=0;
int main(){
    cin>>n>>d>>a;
    for(int i=1;i<=n;i++){
        cin>>p[i].first>>p[i].second;
        p[i].second=(p[i].second+a-1)/a;
    }
    sort(p+1,p+1+n);
    ll pos=1,mx=p[pos].second;ans+=mx;
    for(int i=2;i<=n;i++){
        while(p[i].first-p[pos].first>2*d){
            mx-=p[pos].second;
            pos++;
        }
        if(p[i].second<=mx) p[i].second=0;
        else p[i].second-=mx,mx+=p[i].second;
        ans+=p[i].second;
    }
    cout<<ans<<endl;
}                     
```