#include <stdio.h>
#include <algorithm>
#include <tuple>
#include <queue>
using namespace std;
using ll = long long;

using pii = pair<ll, ll>;
ll n, d, a, x[200010], h[200010];
pii s[200010];

queue<pii> q;
int main() {
	scanf("%lld %lld %lld", &n, &d, &a);
	for (int i = 0; i < n; i++)
		scanf("%lld %lld", x + i, h + i);

	for (int i = 0; i < n; i++)
		s[i] = pii(x[i], h[i]);
	sort(s, s + n);
	for (int i = 0; i < n; i++)
		tie(x[i], h[i]) = s[i];

	ll sum = 0, c = 0;
	for (int i = 0; i < n; i++) {
		while (!q.empty() && q.front().first < x[i])
			sum -= a * q.front().second, q.pop();
		ll u = max(0ll, (h[i] - sum + a - 1) / a);
		sum += a * u;
		c += u;
		q.emplace(x[i] + 2 * d, u);
	}
	printf("%lld\n", c);
}
### 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 <stdio.h>
#include <algorithm>
#include <tuple>
#include <queue>
using namespace std;
using ll = long long;

using pii = pair<ll, ll>;
ll n, d, a, x[200010], h[200010];
pii s[200010];

queue<pii> q;
int main() {
	scanf("%lld %lld %lld", &n, &d, &a);
	for (int i = 0; i < n; i++)
		scanf("%lld %lld", x + i, h + i);

	for (int i = 0; i < n; i++)
		s[i] = pii(x[i], h[i]);
	sort(s, s + n);
	for (int i = 0; i < n; i++)
		tie(x[i], h[i]) = s[i];

	ll sum = 0, c = 0;
	for (int i = 0; i < n; i++) {
		while (!q.empty() && q.front().first < x[i])
			sum -= a * q.front().second, q.pop();
		ll u = max(0ll, (h[i] - sum + a - 1) / a);
		sum += a * u;
		c += u;
		q.emplace(x[i] + 2 * d, u);
	}
	printf("%lld\n", c);
}
```