#include <iostream>
#include <cstdio>
#include <algorithm>
#define F first
#define S second
using namespace std;
typedef long long ll;

ll n, d, a, pos, k, ans, cnt[200005];
pair<ll, ll> b[200005];

int main() {
	ll i, j;
	cin >> n >> d >> a;
	for(i=0; i<n; i++) scanf("%lld%lld", &b[i].F, &b[i].S);
	sort(b, b+n);
	for(i=j=0; i<n; i++) {
		for(; j<n && b[j].F <= b[i].F + 2*d; j++);
		j--;
		k = max(0LL, (b[i].S - cnt[i]*a + a-1) / a);
		ans += k;
		cnt[i] += k;
		cnt[j+1] -= k;
		cnt[i+1] += cnt[i];
	}
	cout << ans;
	return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <iostream>
#include <cstdio>
#include <algorithm>
#define F first
#define S second
using namespace std;
typedef long long ll;

ll n, d, a, pos, k, ans, cnt[200005];
pair<ll, ll> b[200005];

int main() {
	ll i, j;
	cin >> n >> d >> a;
	for(i=0; i<n; i++) scanf("%lld%lld", &b[i].F, &b[i].S);
	sort(b, b+n);
	for(i=j=0; i<n; i++) {
		for(; j<n && b[j].F <= b[i].F + 2*d; j++);
		j--;
		k = max(0LL, (b[i].S - cnt[i]*a + a-1) / a);
		ans += k;
		cnt[i] += k;
		cnt[j+1] -= k;
		cnt[i+1] += cnt[i];
	}
	cout << ans;
	return 0;
}

```