#include<bits/stdc++.h>
using namespace std;
#define int long long

int N,D,A;
pair<int, int> m[200000];

signed main() {
  cin >> N >> D >> A;
  for(int i=0; i<N; i++) cin >> m[i].first >> m[i].second;

  sort(m, m+N);

  int ans = 0;
  int damage = 0;
  queue<pair<int, int>> q;

  for(int i=0; i<N; i++) {
    int Xi=m[i].first;
    int Hi=m[i].second;

    while (q.size() && q.front().first < Xi) {damage -= q.front().second; q.pop();}

    Hi -= damage;
    if(Hi <= 0) continue;
    int a = (Hi+A-1) / A;
    ans += a;

    damage += a*A;
    q.emplace(Xi + 2*D, a*A);
  }
  cout << ans << endl;
}

### 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<bits/stdc++.h>
using namespace std;
#define int long long

int N,D,A;
pair<int, int> m[200000];

signed main() {
  cin >> N >> D >> A;
  for(int i=0; i<N; i++) cin >> m[i].first >> m[i].second;

  sort(m, m+N);

  int ans = 0;
  int damage = 0;
  queue<pair<int, int>> q;

  for(int i=0; i<N; i++) {
    int Xi=m[i].first;
    int Hi=m[i].second;

    while (q.size() && q.front().first < Xi) {damage -= q.front().second; q.pop();}

    Hi -= damage;
    if(Hi <= 0) continue;
    int a = (Hi+A-1) / A;
    ans += a;

    damage += a*A;
    q.emplace(Xi + 2*D, a*A);
  }
  cout << ans << endl;
}

```