#include <bits/stdc++.h>
using namespace std;
long long int sum[200005];
int main(){
    long long int n,d,a;
    cin >> n >> d >> a;
    pair<int,long long int> p[n];
    for(int i=0;i<n;i++){
        cin >> p[i].first >> p[i].second;
        p[i].second = (p[i].second+a-1)/a;
    }
    sort(p,p+n);
    int r[n];
    int tmp=n-1;
    for(int i=n-1;i>=0;i--){
        while(p[tmp].first+2*d>=p[i].first && tmp>=0)tmp--;
        r[i]=tmp+1;
    }
    long long int ans=0;
    for(int i=0;i<n;i++){
        p[i].second-=(sum[i]-sum[r[i]]);
        sum[i+1]=sum[i]+max(0LL,p[i].second);
        ans+=max(0LL,p[i].second);
    }
    cout << ans << endl;
}
### 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;
long long int sum[200005];
int main(){
    long long int n,d,a;
    cin >> n >> d >> a;
    pair<int,long long int> p[n];
    for(int i=0;i<n;i++){
        cin >> p[i].first >> p[i].second;
        p[i].second = (p[i].second+a-1)/a;
    }
    sort(p,p+n);
    int r[n];
    int tmp=n-1;
    for(int i=n-1;i>=0;i--){
        while(p[tmp].first+2*d>=p[i].first && tmp>=0)tmp--;
        r[i]=tmp+1;
    }
    long long int ans=0;
    for(int i=0;i<n;i++){
        p[i].second-=(sum[i]-sum[r[i]]);
        sum[i+1]=sum[i]+max(0LL,p[i].second);
        ans+=max(0LL,p[i].second);
    }
    cout << ans << endl;
}
```