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

Solution (in JAVA):

Solution:


import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		long d = sc.nextInt();
		long a = sc.nextInt();

		Fox foxes[] = new Fox[n];
		for(int i = 0; i < n; i++){
			foxes[i] = new Fox(sc.nextInt(), sc.nextInt());
		}

		Arrays.sort(foxes);

		SqrtDecomposition sq = new SqrtDecomposition(n);

		for(int i = 0; i < n; i++){
			sq.add(i + 1, i + 2, foxes[i].hp);
		}

		long count = 0;
		int j = 1;
		for(int i = 0; i < n; i++){
			if(sq.get(i + 1) <= 0){
//				System.out.println(sq.get(i + 1));
//				j = Math.max(j, i + 2);
				continue;
			}
			while(j < n && foxes[j].pos <= foxes[i].pos + 2*d){
				j++;
			}

			long tmpNum = (sq.get(i + 1) + a - 1)/a;
			count += tmpNum;
//			System.out.println("bomb : "+i+" (pos "+foxes[i].pos + ") "+(j - 1)+" (pos "+foxes[j - 1].pos+") " + tmpNum +" times");
			sq.add(i + 1, j + 1, -a*tmpNum);
		}

		System.out.println(count);
	}

}


class SqrtDecomposition {
	int N, K;
	long[] data;
	long[] bucketAdd;
	static final int sqrtN = 512;

	public SqrtDecomposition(int n) {
		N = n;
		K = (N + sqrtN - 1) / sqrtN;
		data = new long[K * sqrtN];
		bucketAdd = new long[K];
	}

	void add(int s, int t, long x) {
		for (int k = 0; k < K; ++k) {
			int l = k * sqrtN, r = (k + 1) * sqrtN;
			if (r <= s || t <= l)
				continue;
			if (s <= l && r <= t) {
				bucketAdd[k] += x;
			} else {
				for (int i = Math.max(s, l); i < Math.min(t, r); ++i) {
					data[i] += x;
				}
			}
		}
	}

	long get(int s) {
		int k = s / sqrtN;
		return data[s] + bucketAdd[k];
	}
}

class Fox implements Comparable<Fox> {
	long pos;
	long hp;

	Fox(int pos, int hp){
		this.pos = pos;
		this.hp = hp;
	}

	@Override
	public int compareTo(Fox f) {
		// TODO 自動生成されたメソッド・スタブ
		return (int)(this.pos - f.pos);
	}
}