#include<bits/stdc++.h>
using namespace std;
long long a,b,c[200005],n,d,ans,l,k;
struct node
{
    long long x,y;
} w[200005];
inline bool cmp(node x,node y)
{
    return x.x<y.x;
}
int main()
{
    scanf("%lld%lld%lld",&n,&d,&a);
    d*=2;
    for(int i=1; i<=n; ++i)
    {
        scanf("%lld%lld",&w[i].x,&w[i].y);
        w[i].y=(w[i].y+a-1)/a;
    }
    sort(w+1,w+n+1,cmp);
    l=1;
    for(int i=1; i<=n; ++i)
    {
        while(l<i&&w[i].x-w[l].x>d)
        {
            k-=c[l++];
        }
        c[i]=w[i].y-k;
        if(c[i]<0)c[i]=0;
        ans+=c[i];
        k+=c[i];
    }
    printf("%lld",ans);
}

### 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;
long long a,b,c[200005],n,d,ans,l,k;
struct node
{
    long long x,y;
} w[200005];
inline bool cmp(node x,node y)
{
    return x.x<y.x;
}
int main()
{
    scanf("%lld%lld%lld",&n,&d,&a);
    d*=2;
    for(int i=1; i<=n; ++i)
    {
        scanf("%lld%lld",&w[i].x,&w[i].y);
        w[i].y=(w[i].y+a-1)/a;
    }
    sort(w+1,w+n+1,cmp);
    l=1;
    for(int i=1; i<=n; ++i)
    {
        while(l<i&&w[i].x-w[l].x>d)
        {
            k-=c[l++];
        }
        c[i]=w[i].y-k;
        if(c[i]<0)c[i]=0;
        ans+=c[i];
        k+=c[i];
    }
    printf("%lld",ans);
}

```