#include <bits/stdc++.h>
#define ll long long
using namespace std;
pair <ll,ll> p[200005];
ll c[200005];
ll n,d,a;
int main()
{
    cin>>n>>d>>a;
    d*=2;
    ll x,y;
    for(int i=1;i<=n;i++)
    {
        cin>>x>>y;
        y=(y+a-1)/a;
        p[i]={x,y};
    }

    sort(p+1,p+n+1);
    ll k=0,ans=0;
    ll l=1;
    for(int i=1;i<=n;i++)
    {
        while(l<i&&p[i].first-p[l].first>d) k-=c[l++];
        c[i]=p[i].second-k;
        if(c[i]<0) c[i]=0;
        ans+=c[i];
        k+=c[i];
    }
    cout<<ans<<endl;
}

### Prompt

Please formulate 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>
#define ll long long
using namespace std;
pair <ll,ll> p[200005];
ll c[200005];
ll n,d,a;
int main()
{
    cin>>n>>d>>a;
    d*=2;
    ll x,y;
    for(int i=1;i<=n;i++)
    {
        cin>>x>>y;
        y=(y+a-1)/a;
        p[i]={x,y};
    }

    sort(p+1,p+n+1);
    ll k=0,ans=0;
    ll l=1;
    for(int i=1;i<=n;i++)
    {
        while(l<i&&p[i].first-p[l].first>d) k-=c[l++];
        c[i]=p[i].second-k;
        if(c[i]<0) c[i]=0;
        ans+=c[i];
        k+=c[i];
    }
    cout<<ans<<endl;
}

```