MULTQ3 - Multiples of 3

There are N numbers a[0], a[1] ... a[N - 1]. Initially all are 0. You have to perform two types of operations :

  1. Increase the numbers between indices A and B (inclusive) by 1. This is represented by the command "0 A B"
  2. Answer how many numbers between indices A and B (inclusive) are divisible by 3. This is represented by the command "1 A B".

Input

The first line contains two integers, N and Q. Each of the next Q lines are either of the form "0 A B" or "1 A B" as mentioned above.

Output

Output 1 line for each of the queries of the form "1 A B" containing the required answer for the corresponding query.

Sample

Input:
4 7
1 0 3
0 1 2
0 1 3
1 0 0
0 0 3
1 3 3
1 0 3

Output:
4
1
0
2

Constraints

1 ≤ N ≤ 100000
1 ≤ Q ≤ 100000
0 ≤ A ≤ B ≤ N - 1


Added by:Varun Jalan
Date:2010-09-12
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: NODEJS OBJC VB.NET
Resource:own problem

hide comments
2019-07-10 18:50:22
AC in one go !!!!!!!
2019-06-27 18:35:36
i didn't use bit operations and still got AC :))

Last edit: 2019-06-27 18:35:46
2019-06-27 16:31:54
if u got TLE, let use bit instead of normal operation.... use k >> 1 instead of k * 2

Last edit: 2019-06-27 16:34:43
2019-05-30 15:40:15
Use static array instead of vector and optimize for not visiting unnecessary nodes while update and query.
Also use Fast IO.
2019-04-03 23:10:06
AC after trying about a week,,
Nice problem,,
2019-03-30 07:39:58
woho 1st ac with lazy approach
2019-03-01 21:59:07
AC IN ONE GO !! But nothing really to brag about
2019-02-10 00:25:35
Why do setters keep such time limits which give AC or TLE on the basis of fast IO? If NlogN is expected complexity, then poorer algos (N^2 etc) will timeout even if they make the limit a little less stringent.
2019-02-07 08:47:47
Tle in java, used fast i/o, any other way to resolve it?
2018-11-03 14:37:39
no need prinf and I make three mod =))
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