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 PYTHON):

Solution:
# -*- coding: utf-8 -*-
import sys
from bisect import bisect_left, bisect_right

N,D,A=map(int, sys.stdin.readline().split())
XH=[ map(int, sys.stdin.readline().split()) for _ in range(N) ]
XH.sort()

bit=[ 0 for _ in range(N+1) ]

#BIT 関数は全て1-indexed
def add(a,w):
    while a<=N:
        bit[a]+=w
        a+=a&-a

def sum(a):
    ret=0
    while 0<a:
        ret+=bit[a]
        a-=a&-a
    return ret

#区間[l,r]に一律wを加える
def range_add(l,r,w):
    add(l,w)
    add(r+1,w*-1)

#a番目のモンスターの体力を取得
def get_value(a):
    return sum(a)

X=[float("-inf")]
H=[float("-inf")]
for x,h in XH:
    X.append(x)
    H.append(h)

for i in range(1,N+1):
    if i==1:
        add(i,H[i])
    else:
        add(i,H[i]-H[i-1])  #区間に対する更新ができるBITを作成するために、値の差分をBITに持つ
    
ans=0

for i in range(1,N+1):
    h=get_value(i)
    if h<=0:    #モンスターの体力がゼロ以下だったら何もしない
        pass
    else:
        x=X[i]
        pos=bisect_right(X,x+2*D)-1 #攻撃範囲の一番右側を二分探索で決定
        if h%A==0:   #hがAで割り切れる場合
            cnt=h/A
            range_add(i,pos,A*cnt*-1)
        else:
            cnt=h/A+1   #hをAで割って余りが出る場合は攻撃回数を+１する
            range_add(i,pos,A*cnt*-1)
        ans+=cnt

print ans