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

int main()
{
  long long n, d, a;
  cin >> n >> d >> a;
  vector< pair<int,int> > xh(n);
  vector<long long> dm(n+1);
  for(int i=0; i<(n); ++i){
    cin >> xh[i].first >> xh[i].second;
    dm[i]=0;
  }
  dm[n]=0;
  sort(xh.begin(),xh.end());
  d += d;
  int ix = 0;
  long long dmg = 0, tot = 0;
  for(int i=0; i<(n); ++i){
    while(ix<n && xh[ix].first <= xh[i].first + d) ix++;
    dmg += dm[i];
    if( xh[i].second - dmg > 0 ){
      long long cnt1 = ( xh[i].second - dmg + (a-1) ) / a;
      tot += cnt1;
      cnt1 *= a;
      dmg += cnt1;
      dm[ix] -= cnt1;
    }
  }
  cout << tot << "\n";
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;

int main()
{
  long long n, d, a;
  cin >> n >> d >> a;
  vector< pair<int,int> > xh(n);
  vector<long long> dm(n+1);
  for(int i=0; i<(n); ++i){
    cin >> xh[i].first >> xh[i].second;
    dm[i]=0;
  }
  dm[n]=0;
  sort(xh.begin(),xh.end());
  d += d;
  int ix = 0;
  long long dmg = 0, tot = 0;
  for(int i=0; i<(n); ++i){
    while(ix<n && xh[ix].first <= xh[i].first + d) ix++;
    dmg += dm[i];
    if( xh[i].second - dmg > 0 ){
      long long cnt1 = ( xh[i].second - dmg + (a-1) ) / a;
      tot += cnt1;
      cnt1 *= a;
      dmg += cnt1;
      dm[ix] -= cnt1;
    }
  }
  cout << tot << "\n";
  return 0;
}

```