#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
typedef pair<int,int> P;
int main(){
    int n,d,A,X[200001],H[200001];
    cin>>n>>d>>A;
    rep(i,n) cin>>X[i]>>H[i];
    vector<P> B;
    rep(i,n) B.push_back(P(X[i],H[i]));
    sort(B.begin(),B.end());
    rep(i,n) X[i]=B[i].first,H[i]=B[i].second;
    ll ans[200001],k=0,a=0;
    rep(i,n){
        if(X[i]<2*d) ans[i]=max(0,int(ceil(double(H[i])/A)-k)),k+=ans[i];
        else{
            for(int j=a;j<n&&X[j]<X[i]-2*d;j++) k-=ans[j],a++;
            ans[i]=max(0,int(ceil(double(H[i])/A)-k)),k+=ans[i];
        }
    }
    ll ANS=0;
    rep(i,n) ANS+=ans[i];
    cout<<ANS;
}

### Prompt

Your task is to create 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;
#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
typedef pair<int,int> P;
int main(){
    int n,d,A,X[200001],H[200001];
    cin>>n>>d>>A;
    rep(i,n) cin>>X[i]>>H[i];
    vector<P> B;
    rep(i,n) B.push_back(P(X[i],H[i]));
    sort(B.begin(),B.end());
    rep(i,n) X[i]=B[i].first,H[i]=B[i].second;
    ll ans[200001],k=0,a=0;
    rep(i,n){
        if(X[i]<2*d) ans[i]=max(0,int(ceil(double(H[i])/A)-k)),k+=ans[i];
        else{
            for(int j=a;j<n&&X[j]<X[i]-2*d;j++) k-=ans[j],a++;
            ans[i]=max(0,int(ceil(double(H[i])/A)-k)),k+=ans[i];
        }
    }
    ll ANS=0;
    rep(i,n) ANS+=ans[i];
    cout<<ANS;
}

```