Car Repair Shop
Contest: 2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred)
Difficulty rating: 1600
Tags: implementation

Polycarp starts his own business. Tomorrow will be the first working day of his car repair shop. For now the car repair shop is very small and only one car can be repaired at a given time.

Polycarp is good at marketing, so he has already collected n requests from clients. The requests are numbered from 1 to n in order they came.

The i-th request is characterized by two values: si — the day when a client wants to start the repair of his car, di — duration (in days) to repair the car. The days are enumerated from 1, the first day is tomorrow, the second day is the day after tomorrow and so on.

Polycarp is making schedule by processing requests in the order from the first to the n-th request. He schedules the i-th request as follows:

- If the car repair shop is idle for di days starting from si (si, si + 1, ..., si + di - 1), then these days are used to repair a car of the i-th client.
- Otherwise, Polycarp finds the first day x (from 1 and further) that there are di subsequent days when no repair is scheduled starting from x. In other words he chooses the smallest positive x that all days x, x + 1, ..., x + di - 1 are not scheduled for repair of any car. So, the car of the i-th client will be repaired in the range [x, x + di - 1]. It is possible that the day x when repair is scheduled to start will be less than si.

Given n requests, you are asked to help Polycarp schedule all of them according to the rules above.

Input format:
The first line contains integer n (1 ≤ n ≤ 200) — the number of requests from clients.

The following n lines contain requests, one request per line. The i-th request is given as the pair of integers si, di (1 ≤ si ≤ 109, 1 ≤ di ≤ 5·106), where si is the preferred time to start repairing the i-th car, di is the number of days to repair the i-th car.

The requests should be processed in the order they are given in the input.

Output format:
Print n lines. The i-th line should contain two integers — the start day to repair the i-th car and the finish day to repair the i-th car.

Constraints:
Time limit: 2.0 s
Memory limit: 512.0 MB

Example input:
3
9 2
7 3
2 4

Example output:
9 10
1 3
4 7

Example input:
4
1000000000 1000000
1000000000 1000000
100000000 1000000
1000000000 1000000

Example output:
1000000000 1000999999
1 1000000
100000000 100999999
1000001 2000000

Editorial:
Let store all ranges which already occupied for some repair in the vector and add queries in the chronological order. Here we need to say that after each add range in this vector we need to sort elements like a pairs, so all vector elements always arrange in increasing order.
Let a is preferred start time and b is preferred finish time for the i-th request (a = si and b = si + di - 1).
If in the vector there is no range which intersects or touches with the range [a, b] means that i-th repair can be done in preferred time and we need to add range [a, b] in the vector with occupied ranges.
In the other case we need to find the first two adjacent ranges (the first means with lowest indexes) between which the range with length b - a + 1 can be fit, press to the left range the current range with length b - a + 1 as close as possible and add this range in the vector.
Do not forget to add fake range [0, 0] to the vector in the beginning. It makes easier the processing of the case when the repair can be started in the first day.