#include <bits/stdc++.h>
using namespace std;

const long long maxn = 2e5+10, inf = 1e18+10;
long long n, d, a;
pair<long long,long long> p[maxn];
long long tag[maxn];

signed main() {
	cin >> n >> d >> a;
	for (long long i = 0; i < n; ++i) {
		cin >> p[i].first >> p[i].second;
	}
	sort(p, p+n);
	p[n].first = inf;
	long long cur = 0, ans = 0;
	for (long long i = 0; i < n; ++i) {
		cur += tag[i];
//		cout << i << ' ' << cur << ' ' << ans << endl;
		if (cur < p[i].second) {
			long long y = upper_bound(p, p+n, make_pair(p[i].first+2*d, inf))-p, tmp = ((p[i].second-cur+a-1)/a);
			tag[y] -= tmp*a, cur += tmp*a;
			ans += tmp;
		}
	}
	cout << ans << endl;
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;

const long long maxn = 2e5+10, inf = 1e18+10;
long long n, d, a;
pair<long long,long long> p[maxn];
long long tag[maxn];

signed main() {
	cin >> n >> d >> a;
	for (long long i = 0; i < n; ++i) {
		cin >> p[i].first >> p[i].second;
	}
	sort(p, p+n);
	p[n].first = inf;
	long long cur = 0, ans = 0;
	for (long long i = 0; i < n; ++i) {
		cur += tag[i];
//		cout << i << ' ' << cur << ' ' << ans << endl;
		if (cur < p[i].second) {
			long long y = upper_bound(p, p+n, make_pair(p[i].first+2*d, inf))-p, tmp = ((p[i].second-cur+a-1)/a);
			tag[y] -= tmp*a, cur += tmp*a;
			ans += tmp;
		}
	}
	cout << ans << endl;
	return 0;
}

```