Problem:
Score : 600 points

Problem Statement

There are N robots numbered 1 to N placed on a number line. Robot i is placed at coordinate X_i. When activated, it will travel the distance of D_i in the positive direction, and then it will be removed from the number line. All the robots move at the same speed, and their sizes are ignorable.

Takahashi, who is a mischievous boy, can do the following operation any number of times (possibly zero) as long as there is a robot remaining on the number line.

Choose a robot and activate it. This operation cannot be done when there is a robot moving.

While Robot i is moving, if it touches another robot j that is remaining in the range [X_i, X_i + D_i) on the number line, Robot j also gets activated and starts moving. This process is repeated recursively.

How many possible sets of robots remaining on the number line are there after Takahashi does the operation some number of times? Compute this count modulo 998244353, since it can be enormous.

Constraints

1 \leq N \leq 2 \times 10^5

-10^9 \leq X_i \leq 10^9

1 \leq D_i \leq 10^9

X_i \neq X_j (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 D_1
:
X_N D_N

Output

Print the number of possible sets of robots remaining on the number line, modulo 998244353.

Sample Input 1

2
1 5
3 3

Sample Output 1

3

There are three possible sets of robots remaining on the number line: \{1, 2\}, \{1\}, and \{\}.

These can be achieved as follows:

If Takahashi activates nothing, the robots \{1, 2\} will remain.

If Takahashi activates Robot 1, it will activate Robot 2 while moving, after which there will be no robots on the number line. This state can also be reached by activating Robot 2 and then Robot 1.

If Takahashi activates Robot 2 and finishes doing the operation, the robot \{1\} will remain.

Sample Input 2

3
6 5
-1 10
3 3

Sample Output 2

5

There are five possible sets of robots remaining on the number line: \{1, 2, 3\}, \{1, 2\}, \{2\}, \{2, 3\}, and \{\}.

Sample Input 3

4
7 10
-10 3
4 3
-4 3

Sample Output 3

16

None of the robots influences others.

Sample Input 4

20
-8 1
26 4
0 5
9 1
19 4
22 20
28 27
11 8
-3 20
-25 17
10 4
-18 27
24 28
-11 19
2 27
-2 18
-1 12
-24 29
31 29
29 7

Sample Output 4

110

Remember to print the count modulo 998244353.

Solution in LuaJIT (2.0.4) (judge verdict: Accepted)

local mfl, mce = math.floor, math.ceil
local mmi, mma = math.min, math.max
local bls, brs = bit.lshift, bit.rshift
local SegTree = {}
SegTree.updateAll = function(self)
  for i = self.stagenum - 1, 1, -1 do
    local cnt = bls(1, i - 1)
    for j = 1, cnt do
      self.stage[i][j] = self.func(self.stage[i + 1][j * 2 - 1], self.stage[i + 1][j * 2])
    end
  end
end
SegTree.create = function(self, n, func, emptyvalue)
  self.func, self.emptyvalue = func, emptyvalue
  local stagenum, mul = 1, 1
  self.stage = {{}}
  while mul < n do
    mul, stagenum = mul * 2, stagenum + 1
    self.stage[stagenum] = {}
  end
  self.stagenum = stagenum
  for i = 1, mul do self.stage[stagenum][i] = emptyvalue end
  self:updateAll()
end
SegTree.getRange = function(self, left, right)
  if left == right then return self.stage[self.stagenum][left] end
  local stagenum = self.stagenum
  local ret = self.emptyvalue
  while left <= right do
    local stage, sz = 1, bls(1, stagenum - 1)
    local len = right - left + 1
    while (left - 1) % sz ~= 0 or len < sz do
      stage, sz = stage + 1, brs(sz, 1)
    end
    ret = self.func(ret, self.stage[stage][1 + brs(left - 1, stagenum - stage)])
    left = left + sz
  end
  return ret
end
SegTree.setValue = function(self, idx, value, silent)
  self.stage[self.stagenum][idx] = value
  if not silent then
    for i = self.stagenum - 1, 1, -1 do
      local dst = brs(idx + 1, 1)
      local rem = dst * 4 - 1 - idx
      self.stage[i][dst] = self.func(self.stage[i + 1][idx], self.stage[i + 1][rem])
      idx = dst
    end
  end
end
SegTree.new = function(n, func, emptyvalue)
  local obj = {}
  setmetatable(obj, {__index = SegTree})
  obj:create(n, func, emptyvalue)
  return obj
end

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 mod = 998244353
local function badd(x, y) return (x + y) % mod end

local n = io.read("*n")
local t = {}
for i = 1, n do
  t[i] = {}
  t[i][1], t[i][2] = io.read("*n", "*n")
end
table.sort(t, function(a, b) return a[1] < b[1] end)
t[n + 1] = {3 * 1000000007, 0}

local stReach = SegTree.new(n + 1, mma, 0)
for i = 1, n + 1 do
  stReach:setValue(i, i)
end

local dp = {}
for i = 1, n + 1 do
  dp[i] = 0
end
dp[n + 1] = 1
for i = n, 1, -1 do
  local dstLen = lower_bound(t, t[i][1] + t[i][2])
  local dstPos = stReach:getRange(i, dstLen - 1)
  stReach:setValue(i, dstPos)
  dp[i] = badd(dp[dstPos + 1], dp[i + 1])
end
print(dp[1])