#include <bits/stdc++.h>
#define LL long long
#define PII pair<long long,long long>
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int maxn = 2e5+10;
int n;
LL a,d;
PII b[maxn];
LL cnt[maxn];
int main(){
  	cin>>n>>d>>a;
  	d<<=1;
  	for (int i=1;i<=n;i++){
    	cin>>b[i].fi>>b[i].se;
    }
  	sort(b+1,b+1+n);
  	int tail = 1;
  	LL sum = 0;
  	for (int i=1;i<=n;i++){
      	cnt[i] += cnt[i-1];
    	if (cnt[i] * a >= b[i].se) continue;
      	b[i].se -= cnt[i] * a;
      	LL t = (b[i].se + a - 1) / a;
      	sum += t;
      	cnt[i] += t;
      	while (tail <= n && b[tail].fi <= b[i].fi + d) tail++;
      	cnt[tail] -= t;
    }
  	cout<<sum<<endl;
	return 0;
}
### Prompt

Generate 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>
#define LL long long
#define PII pair<long long,long long>
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int maxn = 2e5+10;
int n;
LL a,d;
PII b[maxn];
LL cnt[maxn];
int main(){
  	cin>>n>>d>>a;
  	d<<=1;
  	for (int i=1;i<=n;i++){
    	cin>>b[i].fi>>b[i].se;
    }
  	sort(b+1,b+1+n);
  	int tail = 1;
  	LL sum = 0;
  	for (int i=1;i<=n;i++){
      	cnt[i] += cnt[i-1];
    	if (cnt[i] * a >= b[i].se) continue;
      	b[i].se -= cnt[i] * a;
      	LL t = (b[i].se + a - 1) / a;
      	sum += t;
      	cnt[i] += t;
      	while (tail <= n && b[tail].fi <= b[i].fi + d) tail++;
      	cnt[tail] -= t;
    }
  	cout<<sum<<endl;
	return 0;
}
```