#include <bits/stdc++.h>

using namespace std;
#define int long long
const int MOD = 1000000007;

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, D, A;
	cin >> N >> D >> A;
	vector<pair<int, int> > X(N);
	int res = 0;
	for (int i = 0; i < N; i++) {
		cin >> X[i].first >> X[i].second;
	}
	sort(X.begin(), X.end());
	vector<int> T(N + 1, 0);
	int INF = (int)1 << 60;
	for (int i = 0; i < N; i++) {
		if (i > 0) T[i] += T[i - 1];
		int rem = X[i].second - T[i];
		if (rem > 0) {
			int num = (rem + A - 1) / A;
			int r = upper_bound(X.begin(), X.end(), make_pair(X[i].first + 2 * D, INF)) - X.begin();
			T[i] += num * A;
			T[r] -= num * A;
			res += num;
		}
	}
	cout << res << 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 int long long
const int MOD = 1000000007;

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, D, A;
	cin >> N >> D >> A;
	vector<pair<int, int> > X(N);
	int res = 0;
	for (int i = 0; i < N; i++) {
		cin >> X[i].first >> X[i].second;
	}
	sort(X.begin(), X.end());
	vector<int> T(N + 1, 0);
	int INF = (int)1 << 60;
	for (int i = 0; i < N; i++) {
		if (i > 0) T[i] += T[i - 1];
		int rem = X[i].second - T[i];
		if (rem > 0) {
			int num = (rem + A - 1) / A;
			int r = upper_bound(X.begin(), X.end(), make_pair(X[i].first + 2 * D, INF)) - X.begin();
			T[i] += num * A;
			T[r] -= num * A;
			res += num;
		}
	}
	cout << res << endl;
}
```