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.*;

public class Main {

    static int DAT_SIZE = (1<<24)-1;

    static long[] data = new long[DAT_SIZE];
    static long[] datb = new long[DAT_SIZE];

    static void add(int a, int b, long x, int k, int l, int r) {
        if (a<=l && r<=b) {
            data[k] += x;
        } else if (l<b && a<r) {
            datb[k] += (Math.min(b, r) - Math.max(a, l))*x;
            add(a, b, x, k*2+1, l, (l+r)/2);
            add(a, b, x, k*2+2, (l+r)/2, r);
        }
    }

    static long sum(int a, int b, int k, int l, int r) {
        if (b<=l || r<=a) {
            return 0;
        } else if (a<=l && r<=b) {
            return data[k]*(r-l)+datb[k];
        } else {
            long res = (Math.min(b, r) - Math.max(a, l)) * data[k];
            res += sum(a, b, k*2+1, l, (l+r)/2);
            res += sum(a, b, k*2+2, (l+r)/2, r);
            return res;
        }
    }

    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());
        Long[] X = new Long[N];
        ArrayList<long[]> list = new ArrayList<long[]>();
        for (int i=0;i<N;i++) {
            X[i] = Long.parseLong(sc.next());
            long H = Integer.parseInt(sc.next());
            long[] add = {X[i], (H+A-1)/A};
            list.add(add);
        }
        Collections.sort(list, (o1, o2)->Long.compare(o1[0], o2[0]));
        for (int i=0;i<N;i++) {
            add(i, i+1, list.get(i)[1], 0, 0, N);
        }
        Arrays.sort(X);
        int[] index = new int[N];
        long ans = 0;
        for (int i=0;i<N;i++) {
            index[i] = ~Arrays.binarySearch(X, X[i]+2*D+1, (o1, o2)->o1>=o2?1:-1);
            long cnt = sum(i, i+1, 0, 0, N);
            if (cnt<=0) continue;
            add(i, index[i], -cnt, 0, 0, N);
            ans += cnt;
        }
        // System.out.println(Arrays.toString(X));
        // System.out.println(Arrays.toString(index));
        System.out.println(ans);
    }
}