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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

# 解説放送

def main():
    from collections import deque, namedtuple
    from operator import attrgetter
    import sys
    sys.setrecursionlimit(10 ** 7)
    input = sys.stdin.readline

    MOD = 998244353

    Robot = namedtuple('Robot', 'x d reach ind')

    N = int(input())

    robots = []
    for ind in range(N):
        X, D = map(int, input().split())
        robots.append(Robot(X, D, X + D, ind))

    graph = tuple(set() for _ in range(N))  # DAG, graph[robot.ind]={robot.ind,...}
    to_candidates = deque()
    for robot_from in sorted(robots, key=attrgetter('x'), reverse=True):
        """
        右端に位置するロボットから順に「起点」と見なしていく
        既出のロボットで有向辺の行き先として一度も設定されていないものが
        to_candidatesにx昇順で溜められている
        to_candidatesに含まれるロボットで「起点」から接触範囲内のものがある限り、
        「起点」から有向辺を張り続ける
        to_candidatesの先頭候補(最も左側に位置する)に対して辺を張れない場合、
        それ以降の候補のxはより遠い座標で辺を張れないので
        「起点」から張れる辺がないとわかる
        すべてのロボットは「起点」から辺を張られてto_candidatesから外れるか
        to_candidatesに残り、グラフ(森)の各木のrootとなる
        """
        while to_candidates:
            robot_to = to_candidates[0]
            if robot_to.x < robot_from.reach:
                graph[robot_from.ind].add(robot_to.ind)
                to_candidates.popleft()
            else:
                break

        to_candidates.appendleft(robot_from)

    def dfs(v):
        """有向グラフで親に遡ることはないので到達フラグ管理は不要"""
        ret = 1
        for u in graph[v]:
            ret = (ret * dfs(u)) % MOD
        return ret + 1

    root_robots = to_candidates
    ret = 1
    for root_robot in root_robots:
        ret = (ret * dfs(root_robot.ind)) % MOD
    print(ret)
    return


if __name__ == '__main__':
    main()