#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
pii p[200005];
queue<pii> seg;
int main(){
    int n, d, a, x, h;
    scanf("%d%d%d", &n, &d, &a);
    for(int i=1;i<=n;i++) scanf("%d%d", &x, &h), p[i] = pii(x, h);
    sort(p+1, p+1+n);
    LL sum = 0, ans = 0;
    for(int i=1;i<=n;i++){
        x = p[i].first, h = p[i].second;
        while(!seg.empty() && seg.front().first < x){
            sum -= seg.front().second*1ll*a;
            seg.pop();
        }
        if(sum >= h) continue;
        int dv = (h-sum+a-1)/a;
        sum += dv*1ll*a;
        ans += dv;
        seg.push(pii(min((LL)2e9, x+2*1ll*d), dv));
    }
    printf("%lld\n", 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;
typedef long long LL;
typedef pair<int, int> pii;
pii p[200005];
queue<pii> seg;
int main(){
    int n, d, a, x, h;
    scanf("%d%d%d", &n, &d, &a);
    for(int i=1;i<=n;i++) scanf("%d%d", &x, &h), p[i] = pii(x, h);
    sort(p+1, p+1+n);
    LL sum = 0, ans = 0;
    for(int i=1;i<=n;i++){
        x = p[i].first, h = p[i].second;
        while(!seg.empty() && seg.front().first < x){
            sum -= seg.front().second*1ll*a;
            seg.pop();
        }
        if(sum >= h) continue;
        int dv = (h-sum+a-1)/a;
        sum += dv*1ll*a;
        ans += dv;
        seg.push(pii(min((LL)2e9, x+2*1ll*d), dv));
    }
    printf("%lld\n", ans);
}
```