#include <algorithm>
#include <iostream>
#include <queue>
#include <utility>

using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;

ll n, d, a, ans;
P xh[220000];
queue<P> q;

int main() {
  cin >> n >> d >> a;
  
  for (ll i = 0; i < n; i++) {
    cin >> xh[i].first >> xh[i].second;
    xh[i].second = (xh[i].second+a-1)/a;
  }
  
  sort(xh, xh+n);
  
  ll sum = 0;
  for (ll i = 0; i < n; i++) {
    while (q.size() and q.front().first < xh[i].first) {
      sum -= q.front().second;
      q.pop();
    }
    ll num = max(0LL, xh[i].second - sum);
    ans += num;
    sum += num;
    q.push(P(xh[i].first + 2*d, num));
  }
  
  cout << ans << endl;
  return 0;
}
### Prompt

Your task is to create 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 <algorithm>
#include <iostream>
#include <queue>
#include <utility>

using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;

ll n, d, a, ans;
P xh[220000];
queue<P> q;

int main() {
  cin >> n >> d >> a;
  
  for (ll i = 0; i < n; i++) {
    cin >> xh[i].first >> xh[i].second;
    xh[i].second = (xh[i].second+a-1)/a;
  }
  
  sort(xh, xh+n);
  
  ll sum = 0;
  for (ll i = 0; i < n; i++) {
    while (q.size() and q.front().first < xh[i].first) {
      sum -= q.front().second;
      q.pop();
    }
    ll num = max(0LL, xh[i].second - sum);
    ans += num;
    sum += num;
    q.push(P(xh[i].first + 2*d, num));
  }
  
  cout << ans << endl;
  return 0;
}
```