#include<bits/stdc++.h>
typedef  long long ll;
using namespace std;
const int maxn = 2e5 + 100;
const int inf = 2e9;
int a[maxn];
int main() {
	map<int, int>m;
	ios::sync_with_stdio(0);
	ll n, D, A;
	cin >> n >> D >> A;
	for (int i = 1, x; i <= n; i++)
		cin >> a[i] >> x, m[a[i]] = x;
	sort(a + 1, a + 1 + n);
	ll ans = 0, j = 1;
	for (int i = 1; i <= n; i++) {
		int Max = a[i] + 2 * D;
		while (j < n && a[j + 1] <= Max) {
			m[a[j + 1]] += ans * A;
			j++;
		}
		m[a[i]] -= ans * A;
		ll num = m[a[i]] / A + (m[a[i]] % A == 0 ? 0 : 1);
		if (m[a[i]] > 0)
			ans += num;
	}
	cout << ans << endl;
	return 0;
}

### Prompt

Please formulate 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>
typedef  long long ll;
using namespace std;
const int maxn = 2e5 + 100;
const int inf = 2e9;
int a[maxn];
int main() {
	map<int, int>m;
	ios::sync_with_stdio(0);
	ll n, D, A;
	cin >> n >> D >> A;
	for (int i = 1, x; i <= n; i++)
		cin >> a[i] >> x, m[a[i]] = x;
	sort(a + 1, a + 1 + n);
	ll ans = 0, j = 1;
	for (int i = 1; i <= n; i++) {
		int Max = a[i] + 2 * D;
		while (j < n && a[j + 1] <= Max) {
			m[a[j + 1]] += ans * A;
			j++;
		}
		m[a[i]] -= ans * A;
		ll num = m[a[i]] / A + (m[a[i]] % A == 0 ? 0 : 1);
		if (m[a[i]] > 0)
			ans += num;
	}
	cout << ans << endl;
	return 0;
}

```