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 static java.util.Comparator.comparing;

import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;

public class Main {
    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        Solver solver = new Solver(System.in, out);
        solver.solve();
        out.close();
    }
}
class Solver {
	Scanner sc;
	PrintWriter out;
    public Solver(InputStream in, PrintWriter out) {
    	sc = new Scanner(in);
    	this.out = out;
    }
    // ==========================================================
    List<PPL> list = new ArrayList<>();
	Queue<PPL> que = new ArrayDeque<>();
    public void solve() {
    	int N = ni();
    	long D = nl();
    	long A = nl();
    	long x, h;
    	for (int i = 0; i < N; i++) {
			x = nl();
			h = nl();
			list.add(new PPL(x,h));
		}
    	list.sort(comparing(PPL::getKey));
    	long wk, cnt, total = 0, ans = 0;
    	PPL bb;
    	for (PPL pp : list) {
    		wk = pp.val;
    		while(!que.isEmpty() && que.peek().key < pp.key) {
   				bb = que.poll();
   				total -= bb.val;
    		}
    		wk -= total;
    		if(wk > 0) {
   				cnt = (wk + A - 1) / A;
   				ans += cnt;
   				total += A * cnt;
   				que.add(new PPL(pp.key + 2 * D, A * cnt));
    		}
		}
    	out.println(ans);
    }
    class PPL {
    	public long key, val;
    	public PPL(long key, long val) {
    		this.key = key;
    		this.val = val;
    	}
		public long getKey() {
			return key;
		}
		public void setKey(long key) {
			this.key = key;
		}
		public long getVal() {
			return val;
		}
		public void setVal(long val) {
			this.val = val;
		}
    }
    // ------------------------------------------
    // 入力
    // ------------------------------------------
    public int ni() {
    	return sc.nextInt();
    }
    public long nl() {
    	return sc.nextLong();
    }
    public String ns() {
    	return sc.next();
    }
    public char[] nc() {
    	return sc.next().toCharArray();
    }
    public int[] ndi(int N) {
    	int[] ans = new int[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ni();
		}
    	return ans;
    }
    public long[] ndl(int N) {
    	long[] ans = new long[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = nl();
		}
    	return ans;
    }
    public String[] nds(int N) {
    	String[] ans = new String[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ns();
		}
    	return ans;
    }
    public char[][] ndc(int N) {
    	char[][] ans = new char[N][];
    	for (int i = 0; i < N; i++) {
			ans[i] = nc();
		}
    	return ans;
    }
    public int[][] nddi(int N, int S) {
    	int[][] ans = new int[N][S];
    	for (int i = 0; i < N; i++) {
			for (int j = 0; j < S; j++) {
				ans[i][j] = ni();
			}
		}
    	return ans;
    }
}