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

#define x first
#define h second

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  ll n, w, dmg;
  cin >> n >> w >> dmg;
  w = w*2+1;
  vector<pair<ll,ll>> p;
  for (int i = 0; i < n; i++) {
    ll x, h;
    cin >> x >> h;
    p.emplace_back(x, (h+dmg-1)/dmg);
  }
  sort(p.begin(), p.end());
  vector<ll> will(n+1);
  int j = 0;
  ll acc = 0;
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    while (j < n && p[j].x-p[i].x < w) j++;
    acc += will[i];
    ll v = max(p[i].h-acc, 0ll);
    ans += v;
    acc += v;
    will[j] -= v;
  }
  cout << ans << endl;
}

### 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;
typedef long long ll;

#define x first
#define h second

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  ll n, w, dmg;
  cin >> n >> w >> dmg;
  w = w*2+1;
  vector<pair<ll,ll>> p;
  for (int i = 0; i < n; i++) {
    ll x, h;
    cin >> x >> h;
    p.emplace_back(x, (h+dmg-1)/dmg);
  }
  sort(p.begin(), p.end());
  vector<ll> will(n+1);
  int j = 0;
  ll acc = 0;
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    while (j < n && p[j].x-p[i].x < w) j++;
    acc += will[i];
    ll v = max(p[i].h-acc, 0ll);
    ans += v;
    acc += v;
    will[j] -= v;
  }
  cout << ans << endl;
}

```