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


hide comments
ajay_5097: 2016-05-23 19:38:37

IT felt so good !
AC in 1 go :)
Segmented Tree + Lazy Propagation
O (n + Q*logn)

epsilon: 2015-08-26 13:48:38

thanks to vimarsh for fast i/o

karthik1997: 2015-08-09 01:01:10

o(n) algo +fast i/o needed

shakaal: 2015-01-12 22:03:25

truly teaches u the concept fastIO,lazy,segment tree and finally AC

Eddy Cael: 2015-01-07 21:02:11

No necesary fast I/O. I got AC with ordinary Segment Tree + Lazy + cin/cout in 4.60 s.
Edit.- I got AC with Segment Tree + Lazy + printf/scanf in 4.86 s.

Last edit: 2015-01-07 21:04:30
Deepanker Aggarwal: 2014-12-31 01:40:47

Time Limit for languages like Java should be increased
--Edit--
Journey to AC
1) Shifted from Java to C++. (TLE)
2) Added fast I/O (TLE)
3) Changed functions to inline(TLE)
4) Used static initialization of variables (AC) :)

Last edit: 2014-12-31 20:14:01
Vaibhav Gosain: 2014-12-24 09:08:32

time limit is a little too tight ;-)

anonymous: 2014-07-03 01:14:13

same code AC in C++ gives tle in c...

GURVINDER SINGH: 2014-05-28 16:03:23

AC in first attempt no need of fast i/o


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