#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;

int N, D, A;
pair<int, int> M[200200];

int main()
{
	scanf ("%d %d %d", &N, &D, &A);
	for (int i = 0; i < N; i++){
		int x, h; scanf ("%d %d", &x, &h);
		x -= D;
		M[i] = { x,h };
	}
	sort(M, M + N);

	long long drops = 0, have = 0;
	queue<pair<int, long long> > Q;
	for (int i = 0; i < N; i++){
		while (!Q.empty() && Q.front().first < M[i].first){
			have -= Q.front().second;
			Q.pop();
		}
		
		long long nh = M[i].second - have * A;
		if (nh > 0){
			long long d = (nh + A - 1) / A;
			Q.push({ M[i].first + 2 * D, d });
			drops += d;  have += d;
		}
	}

	printf ("%lld\n", drops);

	return 0;
}

### Prompt

In CPP, your task is to solve 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 <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;

int N, D, A;
pair<int, int> M[200200];

int main()
{
	scanf ("%d %d %d", &N, &D, &A);
	for (int i = 0; i < N; i++){
		int x, h; scanf ("%d %d", &x, &h);
		x -= D;
		M[i] = { x,h };
	}
	sort(M, M + N);

	long long drops = 0, have = 0;
	queue<pair<int, long long> > Q;
	for (int i = 0; i < N; i++){
		while (!Q.empty() && Q.front().first < M[i].first){
			have -= Q.front().second;
			Q.pop();
		}
		
		long long nh = M[i].second - have * A;
		if (nh > 0){
			long long d = (nh + A - 1) / A;
			Q.push({ M[i].first + 2 * D, d });
			drops += d;  have += d;
		}
	}

	printf ("%lld\n", drops);

	return 0;
}

```