#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
using namespace std;
const int maxn=200000+66;
struct monster
{
    ll x,h;
}mon[maxn];
bool cmp(monster a,monster b)
{
    return a.x<b.x;
}
ll c[maxn];
int main()
{
    ll n,d,a;
    cin>>n>>d>>a;
    rep(i,1,n)
    {
        cin>>mon[i].x>>mon[i].h;
    }
    sort(mon+1,mon+n+1,cmp);
    ll ans=0;
    //尺取法
    for(int i=1,j=1;i<=n;i++)
    {
        while(j<=n&&mon[j].x<=mon[i].x+2*d)
            j++;//找到不能炸到的
        ll need=max((mon[i].h-c[i]*a+a-1)/a,0ll);
        c[i]+=need;
        c[j]-=need;
        c[i+1]+=c[i];
        ans+=need;
    }
    cout<<ans<<endl;
}

### Prompt

Your task is to create 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 rep(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
using namespace std;
const int maxn=200000+66;
struct monster
{
    ll x,h;
}mon[maxn];
bool cmp(monster a,monster b)
{
    return a.x<b.x;
}
ll c[maxn];
int main()
{
    ll n,d,a;
    cin>>n>>d>>a;
    rep(i,1,n)
    {
        cin>>mon[i].x>>mon[i].h;
    }
    sort(mon+1,mon+n+1,cmp);
    ll ans=0;
    //尺取法
    for(int i=1,j=1;i<=n;i++)
    {
        while(j<=n&&mon[j].x<=mon[i].x+2*d)
            j++;//找到不能炸到的
        ll need=max((mon[i].h-c[i]*a+a-1)/a,0ll);
        c[i]+=need;
        c[j]-=need;
        c[i+1]+=c[i];
        ans+=need;
    }
    cout<<ans<<endl;
}

```