#include<bits/stdc++.h>
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define range(a) (a).begin(),(a).end()
#define print(A,x,n) rep(i,0,n){cout<<(i ? " ":"")<<A[i]x;}cout<<endl;
#define pprint(A,y,m,n) rep(j,0,m){print(A[j],y,n);}
const long mod=1e9+7;
const int size=1e5;
const int inf=1e9;
int main(){
	int N;long D,A;cin>>N>>D>>A;
	pair<long,long> X[N];
	rep(i,0,N) cin>>X[i].first>>X[i].second;
	sort(X,X+N);
	long t[N];long res=0;
	long ans=0;
	int j=0;
	rep(i,0,N){
		while(X[j].first+2*D<X[i].first){
			res-=t[j];
			j++;
		}
        X[i].second-=res*A;
		t[i]=max((X[i].second+A-1)/A,(long)0);
		res+=t[i]; ans+=t[i];
		X[i].second-=A*t[i];
	}
	cout<<ans<<endl;
}
### Prompt

Your challenge is to write 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,x,y) for(int i=x;i<y;i++)
#define range(a) (a).begin(),(a).end()
#define print(A,x,n) rep(i,0,n){cout<<(i ? " ":"")<<A[i]x;}cout<<endl;
#define pprint(A,y,m,n) rep(j,0,m){print(A[j],y,n);}
const long mod=1e9+7;
const int size=1e5;
const int inf=1e9;
int main(){
	int N;long D,A;cin>>N>>D>>A;
	pair<long,long> X[N];
	rep(i,0,N) cin>>X[i].first>>X[i].second;
	sort(X,X+N);
	long t[N];long res=0;
	long ans=0;
	int j=0;
	rep(i,0,N){
		while(X[j].first+2*D<X[i].first){
			res-=t[j];
			j++;
		}
        X[i].second-=res*A;
		t[i]=max((X[i].second+A-1)/A,(long)0);
		res+=t[i]; ans+=t[i];
		X[i].second-=A*t[i];
	}
	cout<<ans<<endl;
}
```