#include <iostream>
#include <cstdio>
#include <algorithm>
#define N 200005
#define X first
#define H second
using namespace std;

typedef long long ll;
ll n, d, a, sum, ans, s[N];
pair<ll, ll> b[N];
int main()
{
	ll i, j;
	cin >> n >> d >> a;
	for (i = 0; i < n; i++) {
		scanf("%lld %lld", &b[i].X, &b[i].H);
	}
	sort(b, b + n);
	for (i = j = 0; i < n; i++) {
		for (; j < i && b[i].X - b[j].X > 2 * d; j++) sum -= s[j];
		s[i] = max(0LL, (b[i].H + a - 1) / a - sum);
		ans += s[i];
		sum += s[i];
	}
    cout << ans << endl;
    return 0;
}

### Prompt

Develop a solution in CPP 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 N 200005
#define X first
#define H second
using namespace std;

typedef long long ll;
ll n, d, a, sum, ans, s[N];
pair<ll, ll> b[N];
int main()
{
	ll i, j;
	cin >> n >> d >> a;
	for (i = 0; i < n; i++) {
		scanf("%lld %lld", &b[i].X, &b[i].H);
	}
	sort(b, b + n);
	for (i = j = 0; i < n; i++) {
		for (; j < i && b[i].X - b[j].X > 2 * d; j++) sum -= s[j];
		s[i] = max(0LL, (b[i].H + a - 1) / a - sum);
		ans += s[i];
		sum += s[i];
	}
    cout << ans << endl;
    return 0;
}

```