#include "bits/stdc++.h"
using namespace std;
using lint = long long;
const int INF = 1e9;
typedef pair<lint, lint> P;

signed main(){
  lint N, D, A; cin >> N >> D >> A;
  vector<P> m(N);
  for(int i = 0; i < N; i++) cin >> m[i].first >> m[i].second;
  sort(m.begin(), m.end());
  queue<P> que;
  lint ans = 0, counter = 0;
  for(int i = 0; i < N; i++){
    while(que.size() && que.front().first < m[i].first){
      counter -= que.front().second;
      que.pop();
    }
    m[i].second -= counter * A;
    if(m[i].second < 0) continue;
    lint num = (m[i].second + A - 1) / A;
    ans += num; counter += num;
    que.push(P(m[i].first + 2 * D, num));
  }
  cout << ans << endl;
}
### 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 "bits/stdc++.h"
using namespace std;
using lint = long long;
const int INF = 1e9;
typedef pair<lint, lint> P;

signed main(){
  lint N, D, A; cin >> N >> D >> A;
  vector<P> m(N);
  for(int i = 0; i < N; i++) cin >> m[i].first >> m[i].second;
  sort(m.begin(), m.end());
  queue<P> que;
  lint ans = 0, counter = 0;
  for(int i = 0; i < N; i++){
    while(que.size() && que.front().first < m[i].first){
      counter -= que.front().second;
      que.pop();
    }
    m[i].second -= counter * A;
    if(m[i].second < 0) continue;
    lint num = (m[i].second + A - 1) / A;
    ans += num; counter += num;
    que.push(P(m[i].first + 2 * D, num));
  }
  cout << ans << endl;
}
```