#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,n) for(long long (i)=0;(i)<(n);++(i))
#define ref(i,a,b) for (long long (i)=(a); (i)<=(b); ++(i))
#define endl '\n'
#define st first
#define nd second
#define pb push_back
#define mp make_pair
const int mx=1e6+5;
ll c[mx];
ll n,d,a;
int main(){
	cin >> n >> d >> a;
	vector < pair < long long ,long long > > v;
	for(int i=0;i<n;++i){
		ll x,y;cin >> x >> y;
		v.pb(mp(x,y));
	}
	long long ans=0;
	sort(v.begin(),v.end());
	ll j=0;
	for(ll i=0;i<n;++i)
	{
		while(j<n and v[j].first<=v[i].first+2*d)
			++j;
		ll need=max((v[i].second-c[i]*a+a-1)/a,0ll);
		ans+=need;
		c[j]=c[j]-need;
		c[i]+=need;
		c[i+1]+=c[i];
	}
	cout << ans << 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;
#define ll long long
#define rep(i,n) for(long long (i)=0;(i)<(n);++(i))
#define ref(i,a,b) for (long long (i)=(a); (i)<=(b); ++(i))
#define endl '\n'
#define st first
#define nd second
#define pb push_back
#define mp make_pair
const int mx=1e6+5;
ll c[mx];
ll n,d,a;
int main(){
	cin >> n >> d >> a;
	vector < pair < long long ,long long > > v;
	for(int i=0;i<n;++i){
		ll x,y;cin >> x >> y;
		v.pb(mp(x,y));
	}
	long long ans=0;
	sort(v.begin(),v.end());
	ll j=0;
	for(ll i=0;i<n;++i)
	{
		while(j<n and v[j].first<=v[i].first+2*d)
			++j;
		ll need=max((v[i].second-c[i]*a+a-1)/a,0ll);
		ans+=need;
		c[j]=c[j]-need;
		c[i]+=need;
		c[i+1]+=c[i];
	}
	cout << ans << endl;
}

```