#include <bits/stdc++.h>
using namespace std;
using ll = long long;
typedef pair<ll, ll> P;

int main(){
	int n; ll D, A; scanf("%d %lld %lld", &n, &D, &A);
	vector<P> a(n);
	for(int i = 0; i < n; i++) scanf("%lld %lld", &a[i].first, &a[i].second);
	sort(a.begin(), a.end());
	vector<ll> x(n), h(n), cnt(n + 1, 0);
	for(int i = 0; i < n; i++) x[i] = a[i].first, h[i] = a[i].second;
	ll ans = 0, id = 0;
	for(int i = 0; i < n; i++){
		if(i != 0) cnt[i] += cnt[i - 1];
		h[i] -= cnt[i] * A;
		if(h[i] <= 0) continue;
		ll num = (h[i] + A - 1) / A;
		ans += num;
		cnt[i + 1] += num;
		while(x[id] - x[i] <= 2 * D && id < n) id++;
		cnt[id] -= num;
	}
	printf("%lld\n", ans);
}
### Prompt

Please formulate 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;
using ll = long long;
typedef pair<ll, ll> P;

int main(){
	int n; ll D, A; scanf("%d %lld %lld", &n, &D, &A);
	vector<P> a(n);
	for(int i = 0; i < n; i++) scanf("%lld %lld", &a[i].first, &a[i].second);
	sort(a.begin(), a.end());
	vector<ll> x(n), h(n), cnt(n + 1, 0);
	for(int i = 0; i < n; i++) x[i] = a[i].first, h[i] = a[i].second;
	ll ans = 0, id = 0;
	for(int i = 0; i < n; i++){
		if(i != 0) cnt[i] += cnt[i - 1];
		h[i] -= cnt[i] * A;
		if(h[i] <= 0) continue;
		ll num = (h[i] + A - 1) / A;
		ans += num;
		cnt[i + 1] += num;
		while(x[id] - x[i] <= 2 * D && id < n) id++;
		cnt[id] -= num;
	}
	printf("%lld\n", ans);
}
```