#include<bits/stdc++.h>
using namespace std; 
typedef long long ll;
#define fi first
#define se second 
#define pb push_back
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n,d,a;
	cin>>n>>d>>a; 
	vector<pair<ll,ll> >v;
	for(int i=0;i<n;i++){
		ll x,h;
		cin>>x>>h;
		v.pb({x,(h+a-1)/a});
	}
	sort(v.begin(),v.end());
	queue<pair<ll,ll> > q;
	ll ans = 0;
	ll cur = 0;
	for(int i=0;i<n;i++){
		while(!q.empty()){
			if(q.front().fi>v[i].fi)break;
			cur-=q.front().se;
			q.pop();
		}
		
		ll trt = max(0ll,v[i].se-cur);
		cur+=trt;
		if(trt)q.push({v[i].fi+d+d+1,trt});
		ans += trt;
	}
	cout<<ans<<endl;
	return 0;
}

### 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; 
typedef long long ll;
#define fi first
#define se second 
#define pb push_back
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n,d,a;
	cin>>n>>d>>a; 
	vector<pair<ll,ll> >v;
	for(int i=0;i<n;i++){
		ll x,h;
		cin>>x>>h;
		v.pb({x,(h+a-1)/a});
	}
	sort(v.begin(),v.end());
	queue<pair<ll,ll> > q;
	ll ans = 0;
	ll cur = 0;
	for(int i=0;i<n;i++){
		while(!q.empty()){
			if(q.front().fi>v[i].fi)break;
			cur-=q.front().se;
			q.pop();
		}
		
		ll trt = max(0ll,v[i].se-cur);
		cur+=trt;
		if(trt)q.push({v[i].fi+d+d+1,trt});
		ans += trt;
	}
	cout<<ans<<endl;
	return 0;
}

```