Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3 2
1 2
5 4
9 2

Sample Output 1

2

First, let us use a bomb at the coordinate 4 to decrease the first and second monsters' health by 2.

Then, use a bomb at the coordinate 6 to decrease the second and third monsters' health by 2.

Now, all the monsters' healths are 0.
We cannot make all the monsters' health drop to 0 or below with just one bomb.

Sample Input 2

9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5

Sample Output 2

5

We should use five bombs at the coordinate 5.

Sample Input 3

3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000

Sample Output 3

3000000000

Watch out for overflow.

Solution in Lua (5.3.2) (judge verdict: Accepted)

local mfl = math.floor
local mce = math.ceil
local function comp(a, b)
  return a[1] < b
end

local function lower_bound(ary, x)
  local num = #ary
  if num == 0 then return 1 end
  if not comp(ary[1], x) then return 1 end
  if comp(ary[num], x) then return num + 1 end
  local min, max = 1, num
  while 1 < max - min do
    local mid = mfl((min + max) / 2)
    if comp(ary[mid], x) then
      min = mid
    else
      max = mid
    end
  end
  return max
end

local n, d, a = io.read("*n", "*n", "*n")
d = d * 2
local t = {}
local imos = {}
for i = 1, n do
  t[i] = {io.read("*n", "*n")}
  imos[i] = 0
end
imos[n + 1] = 0
table.sort(t, function(a, b) return a[1] < b[1] end)
local ret = 0
for i = 1, n do
  local x, h = t[i][1], t[i][2]
  if 1 < i then
    imos[i] = imos[i] + imos[i - 1]
  end
  h = h + imos[i]
  if 0 < h then
    local use = mce(h / a)
    ret = ret + use
    local dst = lower_bound(t, x + d + 1)
    imos[i + 1] = imos[i + 1] - use * a
    imos[dst] = imos[dst] + use * a
  end
end
print(ret)