The pigeonhole principle states that given n+1 pigeons and n holes and that all pigeons are put into a hole, there must exist some hole that contains at least 2 pigeons.
More generally, given p pigeons and n holes, there must exist some hole with ceil(p/n) pigeons.
Guan is not convinced and wants to run a program to prove this.
Each run of the program will simulate p pigeons leaving or entering holes. There will be h holes, numbered from 0 to h-1.
Each pigeon has a number pi and will enter the hole (pi mod h) (i.e. pi%h) when it enters.
Two pigeons may not have the same number.
For each operation, you need to output an integer on each line, the maximum number of pigeons in one hole.
The first line of the input contains two integers, the number of operations n, and the number of holes h.
For the next n lines, each line consists of an integer, either 0 or 1, 1 indicating entry and 0 indicating leaving of a pigeon. This integer is followed by another integer pi, indicating the number of the pigeon entering/leaving.
No pigeon will leave before it has entered.
1<=pi<=1,000,000,000
For each operation, output an integer on each line, the maximum number of pigeons in one hole.
Subtask 1 (30%) n,h<=2000
Subtask 2 (70%) n,h<=2000000
Input
5 3 1 1 1 9001 0 9001 1 2 0 2Output
1 2 1 1 1
Pigeon 9001 shares the same hole as pigeon 1 (hole 1) (9001%3=1), but pigeon 2 does not share the same hole as pigeon 1, hence after operation two, there are two pigeons in the same hole, but not after operation three.
Input
5 3 1 1 1 9001 1 2 0 2 0 9001Output
1 2 2 2 1
| Subtask | Score |
|---|---|
| 1 | 30 |
| 2 | 70 |
| 3 | 0 |