#include<bits/stdc++.h>
using namespace std;
#define int long
using pi=pair<int,int>;
signed main(){
    int n, d, a, res = 0;
    cin >> n >> d >> a;
    vector<pi> enemy(n);
    for (int i = 0; i < n; ++i) cin >> enemy[i].first >> enemy[i].second;
    sort(enemy.begin(), enemy.end());
    int r = 0;
    vector<int> dmg(n,0);
    for (int i = 0; i < n; ++i) {
        if (i) dmg[i] += dmg[i - 1];
        if (enemy[i].second <= dmg[i]) continue;
        while (i > r || (r < n && enemy[i].first + 2 * d >= enemy[r].first)) ++r;
        int hit = (enemy[i].second - dmg[i] + a - 1) / a;
        dmg[i] += hit * a;
        if(r<n)dmg[r] -= hit * a;
        res += hit;
    }
    cout << res;
}
### 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>
using namespace std;
#define int long
using pi=pair<int,int>;
signed main(){
    int n, d, a, res = 0;
    cin >> n >> d >> a;
    vector<pi> enemy(n);
    for (int i = 0; i < n; ++i) cin >> enemy[i].first >> enemy[i].second;
    sort(enemy.begin(), enemy.end());
    int r = 0;
    vector<int> dmg(n,0);
    for (int i = 0; i < n; ++i) {
        if (i) dmg[i] += dmg[i - 1];
        if (enemy[i].second <= dmg[i]) continue;
        while (i > r || (r < n && enemy[i].first + 2 * d >= enemy[r].first)) ++r;
        int hit = (enemy[i].second - dmg[i] + a - 1) / a;
        dmg[i] += hit * a;
        if(r<n)dmg[r] -= hit * a;
        res += hit;
    }
    cout << res;
}
```