#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN=2e5+5;
typedef long long ll;
int n,d,a;
struct node{
	ll x;
	int h;
}mons[MAXN];
bool operator <(node a,node b){
	return a.x<b.x;
}
ll sum[MAXN];
ll ans;
int main(){
	scanf("%d%d%d",&n,&d,&a);
	for(int i=1; i<=n; i++)
		scanf("%lld%d",&mons[i].x,&mons[i].h);
	sort(mons+1,mons+n+1);
	ll cnt=0;
	for(int i=1,r=1; i<=n; i++){
		while(r<n&&mons[r+1].x<=mons[i].x+d*2) r++;
		cnt+=sum[i];
		mons[i].h-=cnt;
		if(mons[i].h<=0) continue;
		ll t=(mons[i].h+a-1)/a;
		ans+=t;
		sum[i]=t*a;
		sum[r+1]-=t*a;
		cnt+=sum[i];
	}
	printf("%lld\n",ans);
	return 0;
}
### 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<cstdio>
#include<algorithm>
using namespace std;
const int MAXN=2e5+5;
typedef long long ll;
int n,d,a;
struct node{
	ll x;
	int h;
}mons[MAXN];
bool operator <(node a,node b){
	return a.x<b.x;
}
ll sum[MAXN];
ll ans;
int main(){
	scanf("%d%d%d",&n,&d,&a);
	for(int i=1; i<=n; i++)
		scanf("%lld%d",&mons[i].x,&mons[i].h);
	sort(mons+1,mons+n+1);
	ll cnt=0;
	for(int i=1,r=1; i<=n; i++){
		while(r<n&&mons[r+1].x<=mons[i].x+d*2) r++;
		cnt+=sum[i];
		mons[i].h-=cnt;
		if(mons[i].h<=0) continue;
		ll t=(mons[i].h+a-1)/a;
		ans+=t;
		sum[i]=t*a;
		sum[r+1]-=t*a;
		cnt+=sum[i];
	}
	printf("%lld\n",ans);
	return 0;
}
```