#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
typedef long long ll;
int cnt(int h, int a) {
  return h % a ? h / a + 1 : h / a;
}
int main() {
  ll n, d, a, sum=0,x=0;
  int h, k=0, l=0;
  cin >> n >> d >> a;
  vector<pair<ll, int>> c(n),v(n);
  rep(i,n)
    cin >> c[i].first >> c[i].second;
  sort(c.begin(), c.end());
  rep(i,n){
    for (int j=k; j<l;++j)
      if(v[j].first<c[i].first){
        k++;
        x-=v[j].second;
      }else break;
    c[i].second-=x;
    if (c[i].second > 0) {
      h = cnt(c[i].second, a);
      sum += h;
      h *= a;
      x+=h;
      v[l].first = c[i].first+2*d;
      v[l].second = h;
      l++;
    }
  }
  cout << sum;
}

### Prompt

In Cpp, your task is to solve 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 rep(i, n) for (int i = 0; i < (int)(n); ++i)
typedef long long ll;
int cnt(int h, int a) {
  return h % a ? h / a + 1 : h / a;
}
int main() {
  ll n, d, a, sum=0,x=0;
  int h, k=0, l=0;
  cin >> n >> d >> a;
  vector<pair<ll, int>> c(n),v(n);
  rep(i,n)
    cin >> c[i].first >> c[i].second;
  sort(c.begin(), c.end());
  rep(i,n){
    for (int j=k; j<l;++j)
      if(v[j].first<c[i].first){
        k++;
        x-=v[j].second;
      }else break;
    c[i].second-=x;
    if (c[i].second > 0) {
      h = cnt(c[i].second, a);
      sum += h;
      h *= a;
      x+=h;
      v[l].first = c[i].first+2*d;
      v[l].second = h;
      l++;
    }
  }
  cout << sum;
}

```