#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

constexpr int MAXN = 200000;

pair<int, int> monster[MAXN + 2];

int main() {
	int N, D, A;
	cin >> N >> D >> A;
	for(int i = 0; i < N; i++)
		cin >> monster[i].first >> monster[i].second;
	sort(monster, monster + N);
	queue<pair<int, int>> atk;
	int d = 2 * D + 1;
	ll ans = 0;
	for(int i = 0, cur = 0; i < N; i++) {
		int x = monster[i].first, h = (monster[i].second + A - 1) / A;
		while(!atk.empty() && x >= atk.front().first) {
			cur -= atk.front().second;
			atk.pop();
		}
		if(cur < h) {
			ans += h - cur;
			atk.push({x + d, h - cur});
			cur = h;
		}
	}
	cout << ans << '\n';
}

### Prompt

Your challenge is to write 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;

typedef long long ll;

constexpr int MAXN = 200000;

pair<int, int> monster[MAXN + 2];

int main() {
	int N, D, A;
	cin >> N >> D >> A;
	for(int i = 0; i < N; i++)
		cin >> monster[i].first >> monster[i].second;
	sort(monster, monster + N);
	queue<pair<int, int>> atk;
	int d = 2 * D + 1;
	ll ans = 0;
	for(int i = 0, cur = 0; i < N; i++) {
		int x = monster[i].first, h = (monster[i].second + A - 1) / A;
		while(!atk.empty() && x >= atk.front().first) {
			cur -= atk.front().second;
			atk.pop();
		}
		if(cur < h) {
			ans += h - cur;
			atk.push({x + d, h - cur});
			cur = h;
		}
	}
	cout << ans << '\n';
}

```