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
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<(n);i++)

int main(int argc, char const *argv[])
{
	int n, d, a; cin >> n >> d >> a;
	vector<pair<int, int>> v(n);
	rep(i, n) {
		int x, h; cin >> x >> h;
		v[i] = make_pair(x, h);
	}
	sort(v.begin(), v.end());

	ll ans = 0;
	ll sum = 0;
	queue<pair<ll, ll>> que; // <x, damage>
	rep(i, n) {
		ll x = v[i].first, h = v[i].second;
		while(not que.empty() and que.front().first < x) {
			sum -= que.front().second;
			que.pop();
		}
		h -= sum;
		if (h <= 0) continue;
		int time = (h + a - 1) / a;
		ans += time;
		sum += a * time;
		que.push(make_pair(x + 2 * d, a * time));
	}

	cout << ans << endl;
	return 0;
}
