#include <bits/stdc++.h>
using namespace std;
int main(){
long N, D, A, i, j, t, X, H, G, M=0, ans=0; cin >> N >> D >> A; vector< pair<long, long> > V;
for(i=0; i<N; i++){ scanf("%ld %ld", &X, &H); G=(H+A-1)/A; M=max(M, G); V.push_back(make_pair(X, G)); }
sort(V.begin(), V.end());
if(V[0].first+2*D>=V[N-1].first){ ans=M; }
else{
for(i=0; i<N; i++){ 
  if(V[i].second<=0){ continue; }
  for(t=V[i].first+2*D, j=i+1; j<N; j++){ if(V[j].first<=t){ V[j].second-=V[i].second; }else{ break; } }
  ans+=V[i].second;
}
}
cout << ans << "\n";
}
### Prompt

Create a solution in CPP for 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;
int main(){
long N, D, A, i, j, t, X, H, G, M=0, ans=0; cin >> N >> D >> A; vector< pair<long, long> > V;
for(i=0; i<N; i++){ scanf("%ld %ld", &X, &H); G=(H+A-1)/A; M=max(M, G); V.push_back(make_pair(X, G)); }
sort(V.begin(), V.end());
if(V[0].first+2*D>=V[N-1].first){ ans=M; }
else{
for(i=0; i<N; i++){ 
  if(V[i].second<=0){ continue; }
  for(t=V[i].first+2*D, j=i+1; j<N; j++){ if(V[j].first<=t){ V[j].second-=V[i].second; }else{ break; } }
  ans+=V[i].second;
}
}
cout << ans << "\n";
}
```