#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 100;
ll c[N];

struct Node {
    int x, h;
    inline bool operator < (const Node& t) const {
        return x < t.x;
    }
} p[N];

int main()
{
    int n, a, d;
    cin >> n >> d >> a;
    for(int i = 1; i <= n; ++i)
        cin >> p[i].x >> p[i].h;
    sort(p + 1, p + n + 1);
    
    ll ans = 0;
    // 双指针
    for(int i = 1, j = 1; i <= n; ++i) {
        while(j <= n && p[j].x <= p[i].x + 2 * d) ++j;
        ll need = max((p[i].h - c[i] * a + a - 1) / a, 0ll);
        ans += need;
        // 构造差分序列
        c[i] += need;
        c[j] -= need;
        c[i + 1] += c[i];
    }
    cout << ans << endl;
    return 0;
}
### 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>
using namespace std;
typedef long long ll;
const int N = 2e5 + 100;
ll c[N];

struct Node {
    int x, h;
    inline bool operator < (const Node& t) const {
        return x < t.x;
    }
} p[N];

int main()
{
    int n, a, d;
    cin >> n >> d >> a;
    for(int i = 1; i <= n; ++i)
        cin >> p[i].x >> p[i].h;
    sort(p + 1, p + n + 1);
    
    ll ans = 0;
    // 双指针
    for(int i = 1, j = 1; i <= n; ++i) {
        while(j <= n && p[j].x <= p[i].x + 2 * d) ++j;
        ll need = max((p[i].h - c[i] * a + a - 1) / a, 0ll);
        ans += need;
        // 构造差分序列
        c[i] += need;
        c[j] -= need;
        c[i + 1] += c[i];
    }
    cout << ans << endl;
    return 0;
}
```