LITE - Light Switching


Farmer John tries to keep the cows sharp by letting them play with intellectual toys. One of the larger toys is the lights in the barn.  Each of the N (2 ≤ N ≤ 100,000) cow stalls conveniently numbered 1..N has a colorful light above it.

At the beginning of the evening, all the lights are off. The cows control the lights with a set of N pushbutton switches that toggle the lights; pushing switch i changes the state of light i from off to on or from on to off.

The cows read and execute a list of M (1 ≤ M ≤ 100,000) operations expressed as one of two integers (0 ≤ operation ≤ 1).

The first kind of operation (denoted by a 0 command) includes two subsequent integers Si and Ei (1 ≤ Si ≤ Ei ≤ N) that indicate a starting switch and ending switch. They execute the operation by pushing each pushbutton from Si through Ei inclusive exactly once.

The second kind of operation (denoted by a 1 command) asks the cows to count how many lights are on in the range given by two integers Si and Ei (1 ≤ Si ≤ Ei ≤ N) which specify the inclusive range in which the cows should count the number of lights that are on.

Help FJ ensure the cows are getting the correct answer by processing the list and producing the proper counts.

Input

Line 1: Two space-separated integers: N and M
Lines 2 to M+1: Each line represents an operation with three space-separated integers: operation, Si, and Ei

Output

Lines 1: number of queries: For each output query, print the count as an integer by itself on a single line.

Example

Input:
4 5
0 1 2
0 2 4
1 2 3
0 2 4
1 1 4

Output:
1
2

hide comments
bad_bat: 2019-10-05 17:58:02

AC in one go!
Piece of cakewalk!

fardin_abir: 2019-09-27 07:34:47

Use scanf,printf....

akshay184: 2019-09-04 22:14:32

Finally accepted

akshay184: 2019-09-04 20:52:11

getting wrong answer at 10 th test case

landofkings: 2019-07-20 13:17:58

no need of any io read , simple cin , cout gives AC but in 0.56 seconds.

saurav_paul: 2019-07-11 21:24:38

Basic Lazy Propagation, NIce problem

saurabh8522: 2019-07-10 16:52:35

AC in one go.

mohit2598: 2019-07-10 16:20:18

Anyone going wrong at 10th test case.. Here is what is happening..
Corner case is when you are going to mark a lazy node again to be lazy.
So, while marking the "child nodes" of a node lazy,, check if they are previously set to lazy.. If yes, then mark them unlazy or 0,, else mark them lazy or 1..

dkkv0000: 2019-05-14 22:17:24

after 3 hrs AC :) my first lazy propagation

amantu_amir: 2019-04-03 23:18:50

AC after trying about a week,,,
Same as "Multiple of 3".


Added by:Iqram Mahmud
Date:2010-09-03
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: OBJC VB.NET
Resource:USACO November 08