#include <bits/stdc++.h>
using namespace std;
using i64 = long long;

int main()
{
  i64 n, d, a;
  cin >> n >> d >> a;
  vector<pair<i64, i64>> mob(n);
  for (i64 i = 0; i < n; i++)
    cin >> mob[i].first >> mob[i].second;
  sort(mob.begin(), mob.end());
  vector<i64> cnt(n);
  i64 ans = 0, at = 0;
  for (i64 i = 0; i < n; i++)
  {
    at -= cnt[i];
    mob[i].second -= at * a;
    i64 tmp = (max(0LL, mob[i].second) + a - 1) / a;
    at += tmp;
    ans += tmp;
    i64 it = upper_bound(mob.begin(), mob.end(), make_pair(mob[i].first + 2 * d, (i64)1e9)) - mob.begin();
    if (it < n)
      cnt[it] += tmp;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 i64 = long long;

int main()
{
  i64 n, d, a;
  cin >> n >> d >> a;
  vector<pair<i64, i64>> mob(n);
  for (i64 i = 0; i < n; i++)
    cin >> mob[i].first >> mob[i].second;
  sort(mob.begin(), mob.end());
  vector<i64> cnt(n);
  i64 ans = 0, at = 0;
  for (i64 i = 0; i < n; i++)
  {
    at -= cnt[i];
    mob[i].second -= at * a;
    i64 tmp = (max(0LL, mob[i].second) + a - 1) / a;
    at += tmp;
    ans += tmp;
    i64 it = upper_bound(mob.begin(), mob.end(), make_pair(mob[i].first + 2 * d, (i64)1e9)) - mob.begin();
    if (it < n)
      cnt[it] += tmp;
  }
  cout << ans << endl;
  return 0;
}

```