
#include<bits/stdc++.h> 
using namespace std; 

int main() 
{ 
    long long int n,d,a;
    cin>>n>>d>>a;
    vector<pair<long long int,long long int> > v(n);
    for(int i=0;i<n;i++){
        cin>>v[i].first>>v[i].second;
    }
    long long int sum=0,res=0;
    
    sort(v.begin(),v.end());
    queue<pair<long long int,long long int> > q;
    for(int i=0;i<n;i++){
        while(q.size()>0 && q.front().first < v[i].first){
           sum-=q.front().second;
           q.pop(); 
        }
        v[i].second-=a*sum;
        if(v[i].second <= 0) continue;
        long long int tmp=(v[i].second +a-1)/a;
        res+=tmp;
        q.push({v[i].first + 2*d,tmp});
        sum+=tmp;
    }
    cout<<res;
	return 0; 
} 

### 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; 

int main() 
{ 
    long long int n,d,a;
    cin>>n>>d>>a;
    vector<pair<long long int,long long int> > v(n);
    for(int i=0;i<n;i++){
        cin>>v[i].first>>v[i].second;
    }
    long long int sum=0,res=0;
    
    sort(v.begin(),v.end());
    queue<pair<long long int,long long int> > q;
    for(int i=0;i<n;i++){
        while(q.size()>0 && q.front().first < v[i].first){
           sum-=q.front().second;
           q.pop(); 
        }
        v[i].second-=a*sum;
        if(v[i].second <= 0) continue;
        long long int tmp=(v[i].second +a-1)/a;
        res+=tmp;
        q.push({v[i].first + 2*d,tmp});
        sum+=tmp;
    }
    cout<<res;
	return 0; 
} 

```