#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int MOD = 1e9+7;
 
int main(){
	
	ll n,d,a; cin>>n>>d>>a;
	d*=2;
	vector<pair<ll,ll>> x(n);
	for(auto &i:x) cin>>i.first>>i.second;
	deque<pair<ll,ll>> damage;
	ll cnt{};
	sort(x.begin(),x.end());
	ll sum{};
	for(int i=0; i<n; i++){
		while(!damage.empty()){
			auto p = damage.front();
			if(x.at(i).first-p.first>d){
				sum -= damage.front().second*a;
				damage.pop_front();
			}
			else break;
		}
		if(x.at(i).second<=sum) continue;
		damage.push_back({x.at(i).first,(x.at(i).second-sum+a-1)/a});
		cnt += (x.at(i).second-sum+a-1)/a;
		sum += (x.at(i).second-sum+a-1)/a*a;
	}
	
	cout << cnt << endl;
}

### Prompt

In CPP, your task is to solve 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;
using ll = long long;
const int MOD = 1e9+7;
 
int main(){
	
	ll n,d,a; cin>>n>>d>>a;
	d*=2;
	vector<pair<ll,ll>> x(n);
	for(auto &i:x) cin>>i.first>>i.second;
	deque<pair<ll,ll>> damage;
	ll cnt{};
	sort(x.begin(),x.end());
	ll sum{};
	for(int i=0; i<n; i++){
		while(!damage.empty()){
			auto p = damage.front();
			if(x.at(i).first-p.first>d){
				sum -= damage.front().second*a;
				damage.pop_front();
			}
			else break;
		}
		if(x.at(i).second<=sum) continue;
		damage.push_back({x.at(i).first,(x.at(i).second-sum+a-1)/a});
		cnt += (x.at(i).second-sum+a-1)/a;
		sum += (x.at(i).second-sum+a-1)/a*a;
	}
	
	cout << cnt << endl;
}

```