#include <bits/stdc++.h>
using namespace std;
#define pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ll long long
#define ld long double
#define all(a) (a).begin(),(a).end()
#define mk make_pair
ll MOD=1000000007;
ll mod=998244353;
int inf=1000001000;
ll INF=1e18+5;


int main() {
  ll n,d,a,u=0,w,ans=0;
  cin >> n >> d >> a;
  vector<pair<ll,ll>> e(n);
  vector<ll> y(n,0);
  rep(i,n) cin >> e[i].first >> e[i].second;
  sort(all(e));
  int k=0;
  rep(i,n){
    w=e[i].first+2*d;
    for (int j=k;j<n;j++){
      if (e[j].first<=w) y[j]=u;
    else break;
    k++;
    }
    ll t=0;
    if (u-y[i]<e[i].second) t=(e[i].second-u+y[i]+a-1)/a;
    u+=a*t;
    ans+=t;
  }
  cout << ans << endl;
}

### Prompt

Please create a solution in Cpp 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 pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ll long long
#define ld long double
#define all(a) (a).begin(),(a).end()
#define mk make_pair
ll MOD=1000000007;
ll mod=998244353;
int inf=1000001000;
ll INF=1e18+5;


int main() {
  ll n,d,a,u=0,w,ans=0;
  cin >> n >> d >> a;
  vector<pair<ll,ll>> e(n);
  vector<ll> y(n,0);
  rep(i,n) cin >> e[i].first >> e[i].second;
  sort(all(e));
  int k=0;
  rep(i,n){
    w=e[i].first+2*d;
    for (int j=k;j<n;j++){
      if (e[j].first<=w) y[j]=u;
    else break;
    k++;
    }
    ll t=0;
    if (u-y[i]<e[i].second) t=(e[i].second-u+y[i]+a-1)/a;
    u+=a*t;
    ans+=t;
  }
  cout << ans << endl;
}

```