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

int main() {
  ll n, d, a;
  cin >> n >> d >> a;
  vector<pair<ll, ll>> v(n);
  for (auto& e : v) cin >> e.first >> e.second;
  sort(v.begin(), v.end());
  vector<ll> x(n), h(n);
  for (int i = 0; i < n; i++) x[i] = v[i].first, h[i] = (v[i].second - 1) / a + 1;

  vector<int> w(n);
  for (int i = 0; i < n; i++) {
    w[i] = upper_bound(x.begin(), x.end(), x[i] + 2 * d) - x.begin();
  }

  ll m = 0, s = 0;
  vector<ll> t(n + 1);
  for (int i = 0; i < n; i++) {
    s += t[i];
    if (s < h[i]) {
      m += h[i] - s;
      t[i] += h[i] - s;
      t[w[i]] -= h[i] - s;
      s = h[i];
    }
  }
  cout << m << endl;
}

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

int main() {
  ll n, d, a;
  cin >> n >> d >> a;
  vector<pair<ll, ll>> v(n);
  for (auto& e : v) cin >> e.first >> e.second;
  sort(v.begin(), v.end());
  vector<ll> x(n), h(n);
  for (int i = 0; i < n; i++) x[i] = v[i].first, h[i] = (v[i].second - 1) / a + 1;

  vector<int> w(n);
  for (int i = 0; i < n; i++) {
    w[i] = upper_bound(x.begin(), x.end(), x[i] + 2 * d) - x.begin();
  }

  ll m = 0, s = 0;
  vector<ll> t(n + 1);
  for (int i = 0; i < n; i++) {
    s += t[i];
    if (s < h[i]) {
      m += h[i] - s;
      t[i] += h[i] - s;
      t[w[i]] -= h[i] - s;
      s = h[i];
    }
  }
  cout << m << endl;
}

```