#include <iostream>
#include <map>

using namespace std;
map<int,int> mp,ind;
typedef long long ll;
ll imos[400010] = {};
int main(){
    ll i,n,d,a;
    cin >> n >> d >> a;
    for(i=0;i<n;i++){
        int x,h; cin >> x >> h;
        mp[x] += h; mp[x + 2*d + 1] += 0;
    }
    int now = 0;
    for(auto x:mp){
        ind[x.first] = now; now++;
    }
    ll ans = 0,dm = 0;
    for(auto x:mp){
        dm += imos[ind[x.first]];
        if(x.second<=dm) continue;
        ll t = (x.second - dm + a - 1)/a;
        ans += t;
        dm += a*t;
        imos[ind[x.first + 2*d + 1]] -= a*t;
    }
    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 <iostream>
#include <map>

using namespace std;
map<int,int> mp,ind;
typedef long long ll;
ll imos[400010] = {};
int main(){
    ll i,n,d,a;
    cin >> n >> d >> a;
    for(i=0;i<n;i++){
        int x,h; cin >> x >> h;
        mp[x] += h; mp[x + 2*d + 1] += 0;
    }
    int now = 0;
    for(auto x:mp){
        ind[x.first] = now; now++;
    }
    ll ans = 0,dm = 0;
    for(auto x:mp){
        dm += imos[ind[x.first]];
        if(x.second<=dm) continue;
        ll t = (x.second - dm + a - 1)/a;
        ans += t;
        dm += a*t;
        imos[ind[x.first + 2*d + 1]] -= a*t;
    }
    cout << ans << endl;
}
```