#define HAVE_STRUCT_TIMESPEC
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200000 + 5;
ll pre[N];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
    cout.tie(NULL);
	ll n,d,a;
	cin >> n >> d >> a;
	vector< pair<ll, ll> > v;
	for(int i = 1; i <= n; ++i){
		ll x,y;
		cin >> x >> y;
		ll z = (y-1) / a + 1;
		v.push_back({x,z});
	}
	sort(v.begin(),v.end());
	ll ans = 0,sum = 0;
	int r = 1;
	for(int i = 0; i < n; ++i){
		while(r < n && v[r].first <= v[i].first + 2 * d)
			++r;
		sum -= pre[i];
		v[i].second -= sum;
		if(v[i].second <= 0)
			continue;
		sum += v[i].second;
		pre[r] += v[i].second;
		ans += v[i].second; 
	}
	cout << ans << endl;
	return 0;
}
### 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
#define HAVE_STRUCT_TIMESPEC
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200000 + 5;
ll pre[N];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
    cout.tie(NULL);
	ll n,d,a;
	cin >> n >> d >> a;
	vector< pair<ll, ll> > v;
	for(int i = 1; i <= n; ++i){
		ll x,y;
		cin >> x >> y;
		ll z = (y-1) / a + 1;
		v.push_back({x,z});
	}
	sort(v.begin(),v.end());
	ll ans = 0,sum = 0;
	int r = 1;
	for(int i = 0; i < n; ++i){
		while(r < n && v[r].first <= v[i].first + 2 * d)
			++r;
		sum -= pre[i];
		v[i].second -= sum;
		if(v[i].second <= 0)
			continue;
		sum += v[i].second;
		pre[r] += v[i].second;
		ans += v[i].second; 
	}
	cout << ans << endl;
	return 0;
}
```