Problem:
Score : 500 points

Problem Statement

You are going to hold a competition of one-to-one game called AtCoder Janken. (Janken is the Japanese name for Rock-paper-scissors.)
N players will participate in this competition, and they are given distinct integers from 1 through N.
The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive).
You cannot assign the same integer to multiple playing fields.
The competition consists of N rounds, each of which proceeds as follows:

For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there.

Then, each player adds 1 to its integer. If it becomes N+1, change it to 1.

You want to ensure that no player fights the same opponent more than once during the N rounds.
Print an assignment of integers to the playing fields satisfying this condition.
It can be proved that such an assignment always exists under the constraints given.

Constraints

1 \leq M

M \times 2 +1 \leq N \leq 200000

Input

Input is given from Standard Input in the following format:

N M

Output

Print M lines in the format below.
The i-th line should contain the two integers a_i and b_i assigned to the i-th playing field.

a_1 b_1
a_2 b_2
:
a_M b_M

Sample Input 1

4 1

Sample Output 1

2 3

Let us call the four players A, B, C, and D, and assume that they are initially given the integers 1, 2, 3, and 4, respectively.

The 1-st round is fought by B and C, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 2, 3, 4, and 1, respectively.

The 2-nd round is fought by A and B, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 3, 4, 1, and 2, respectively.

The 3-rd round is fought by D and A, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 4, 1, 2, and 3, respectively.

The 4-th round is fought by C and D, who has the integers 2 and 3, respectively. After this round, A, B, C, and D have the integers 1, 2, 3, and 4, respectively.

No player fights the same opponent more than once during the four rounds, so this solution will be accepted.

Sample Input 2

7 3

Sample Output 2

1 6
2 5
3 4

Solution in Haskell (GHC 8.8.3) (judge verdict: Wrong Answer)

{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE Rank2Types #-}
{-# LANGUAGE BinaryLiterals #-}
{-# LANGUAGE BangPatterns #-}

-- Default Imports

import Control.Monad
import Control.Monad.ST
import Control.Monad.Trans.Class

import qualified Control.Monad.Reader as Reader
import Control.Monad.Reader (Reader, ReaderT, runReader, runReaderT)

import qualified Data.Array.MArray as MArray
import qualified Data.Array.ST as STArray
import Data.Array.ST (STArray, STUArray, runSTArray, runSTUArray)
import Data.Ix (Ix)

import qualified Data.Bits as Bits
import Data.Bits ((.&.), (.|.))

import qualified Data.ByteString.Char8 as BS
import Data.ByteString.Char8 (ByteString)

import qualified Data.Char as Char
import qualified Data.List as List
import qualified Data.Map.Strict as Map
import Data.Map.Strict (Map)
import qualified Data.Maybe as Maybe
import qualified Data.Set as Set
import Data.Set (Set)

import qualified Data.Vector as Vector
import qualified Data.Vector.Generic as GVector
import qualified Data.Vector.Mutable as MVector
import Data.Vector.Mutable (STVector)
import qualified Data.Vector.Unboxed as UVector

import Numeric

import Debug.Trace
-- ----------

-- Custom Import Here


-- ----------

-- Default Definitions

type Vector = Vector.Vector
-- type GVector = GVector.Vector
type UVector = UVector.Vector

readIntToList :: ByteString -> [Int]
readIntToList = List.unfoldr (fmap (second $ BS.dropWhile Char.isSpace) . BS.readInt)

readToList :: Read a => ByteString -> [a]
readToList = List.unfoldr $ fmap (mapTaken . mapRemain) . span' (not . Char.isSpace)
	where
		mapTaken  = first $ read . BS.unpack
		mapRemain = second $ BS.dropWhile Char.isSpace
		span' pred s
			| BS.null taken = Nothing
			| otherwise     = Just (taken, remain')
			where
				(taken, remain) = BS.span pred s
				remain' = BS.dropWhile Char.isSpace remain

readToListN :: Read a => Int -> ByteString -> [a]
readToListN n = take n . readToList

readIntToVector :: ByteString -> Vector Int
readIntToVector = Vector.unfoldr (fmap stripRemain . BS.readInt)
	where stripRemain = second $ BS.dropWhile Char.isSpace

readIntToUVector :: ByteString -> UVector.Vector Int
readIntToUVector = UVector.unfoldr (fmap stripRemain . BS.readInt)
	where stripRemain = second $ BS.dropWhile Char.isSpace

readIntToVectorN :: Int -> ByteString -> Vector.Vector Int
readIntToVectorN n = Vector.unfoldrN n (fmap stripRemain . BS.readInt)
	where stripRemain = second $ BS.dropWhile Char.isSpace

readIntToUVectorN :: Int -> ByteString -> UVector.Vector Int
readIntToUVectorN n = UVector.unfoldrN n (fmap stripRemain . BS.readInt)
	where stripRemain = second $ BS.dropWhile Char.isSpace


first :: (a -> a') -> (a, b) -> (a', b)
first f (a, b) = (f a, b)

second :: (b -> b') -> (a, b) -> (a, b')
second f (a, b) = (a, f b)

average :: Fractional a => [a] -> a
average xs = sum xs / (fromIntegral . length) xs

-- ----------

calcAs n = [1..(n `div` 2)]
calcBs n = reverse [(n `div` 2 + 1)..n]

showPair (a, b) = do
	putStr . show $ a
	putStr " "
	putStrLn . show $ b

main :: IO ()
main = do
	[n, m] <- readIntToList <$> BS.getLine
	mapM_ showPair $ take m $ zip (calcAs n) (calcBs n)