#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF=ll(1e16)+ll(7);

int main(){
	ll N,D,A;
	cin >> N >> D >> A;
	vector< pair<ll,ll> > XH(N);
	vector<ll> sumd(N+1,0);
	for(int i=0;i<N;i++) cin >> XH[i].first >> XH[i].second;
	sort(XH.begin(),XH.end());
	ll ans = 0;
	for(int i=0;i<N;i++){
		if(i) sumd[i] += sumd[i-1];
		if(sumd[i]>=XH[i].second) continue;
		ll cnt = (XH[i].second-sumd[i]+A-1)/A;
		ans += cnt;
		ll r = upper_bound(XH.begin(),XH.end(),make_pair(XH[i].first+2*D,INF))-XH.begin();
		sumd[i] += cnt*A;
		sumd[r] -= cnt*A;
	}
	cout << ans << endl;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 ll INF=ll(1e16)+ll(7);

int main(){
	ll N,D,A;
	cin >> N >> D >> A;
	vector< pair<ll,ll> > XH(N);
	vector<ll> sumd(N+1,0);
	for(int i=0;i<N;i++) cin >> XH[i].first >> XH[i].second;
	sort(XH.begin(),XH.end());
	ll ans = 0;
	for(int i=0;i<N;i++){
		if(i) sumd[i] += sumd[i-1];
		if(sumd[i]>=XH[i].second) continue;
		ll cnt = (XH[i].second-sumd[i]+A-1)/A;
		ans += cnt;
		ll r = upper_bound(XH.begin(),XH.end(),make_pair(XH[i].first+2*D,INF))-XH.begin();
		sumd[i] += cnt*A;
		sumd[r] -= cnt*A;
	}
	cout << ans << endl;
}

```