#include<bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define len(x) (int)(x.size())
#define all(vs) vs.begin(), vs.end()
using ll = long long;
using ii = pair<ll, ll>;

int32_t main(){
	ios::sync_with_stdio(0);	
	int n;
	ll d, a;
	cin >> n >> d >> a;
	d*=2LL;
	vector<ii> vs(n);
	for(auto &x:vs) cin >> x.ff >> x.ss;
	queue<ii> q;
	ll acc = 0;
	sort(all(vs));
	ll ans = 0;
	for(auto _x:vs){
		int x = _x.ff;
		int h = _x.ss;
		while(not q.empty() and q.front().ff<x) acc-=q.front().ss, q.pop();
		
		h = max(0LL, h-acc);
		ll k = (h+a-1)/a;
		ans += k;
		acc += a*k;
		q.push({x+d, a*k});

	}
	cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution 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;
#define ff first
#define ss second
#define len(x) (int)(x.size())
#define all(vs) vs.begin(), vs.end()
using ll = long long;
using ii = pair<ll, ll>;

int32_t main(){
	ios::sync_with_stdio(0);	
	int n;
	ll d, a;
	cin >> n >> d >> a;
	d*=2LL;
	vector<ii> vs(n);
	for(auto &x:vs) cin >> x.ff >> x.ss;
	queue<ii> q;
	ll acc = 0;
	sort(all(vs));
	ll ans = 0;
	for(auto _x:vs){
		int x = _x.ff;
		int h = _x.ss;
		while(not q.empty() and q.front().ff<x) acc-=q.front().ss, q.pop();
		
		h = max(0LL, h-acc);
		ll k = (h+a-1)/a;
		ans += k;
		acc += a*k;
		q.push({x+d, a*k});

	}
	cout << ans << endl;
}

```