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) {
		Scanner sc = new Scanner(System.in);
		
		int n = Integer.parseInt(sc.next());
		int d = Integer.parseInt(sc.next());
		int a = Integer.parseInt(sc.next());
		
		int [][]x = new int [n][2];
		
		for(int i = 0 ; i < n ;i++) {
			x[i][0] = Integer.parseInt(sc.next());
			x[i][1] = Integer.parseInt(sc.next());
		}
		
		Arrays.sort(x,(p,q) -> Integer.compare(p[0],q[0]));
		
		long []nd = new long [n+1];
		for(int i = 0 ; i < n ;i++) {
			nd[i] = (long)x[i][0];
		}
		nd[n] = 20000000000000000L;
		long []sum = new long [n+1];
		long ans = 0;
		
		for(int i = 0 ; i < n ; i++){
			if(sum[i] < x[i][1]) {
				long A = (x[i][1] - sum[i] + a - 1)/a;	//x[i]を０にするために必要な爆弾の回数
				long D = nd[i] + 2 * d;
				int l = upper_bound(nd,D,n,i);
				sum[i] += A*a;
				sum[l] += -A*a;
				ans += A;
			}
			sum[i+1] += sum[i];
		}
		
		System.out.println(ans);
	}
	
	public static int upper_bound(long[] nd , long d , int n, int f) {		
		int p = n;
		int q = f;
		while(p - q > 1) {
			int mid = (p+q)/2;
			if(nd[mid] == d) {
				return mid+1;
			}
			if(nd[mid] < d) {
				q = mid;
			}
			else {
				p = mid;
			}
		}
		return p;
		
		
	}

}