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: Accepted)


module Main where

import           Control.Monad
import           Debug.Trace
import qualified Data.Vector                   as V
import qualified Data.Vector.Mutable           as VM
import qualified Data.Vector.Unboxed           as VU
import qualified Data.Vector.Unboxed.Mutable   as VUM
import           Data.List
import qualified Data.ByteString               as BS
import qualified Data.ByteString.Char8         as BSC8
import           Data.Char
import           Data.Maybe
import           Data.Ord
--import           Data.Vector.Algorithms.Search
import qualified Data.Array                    as A
import qualified Data.Array.Unboxed            as AU
import qualified Data.Array.IO                 as AIO
import qualified Data.Set                      as S

main = do
    [aN, aM] <- getIL


    let
        ret = if odd aN
            then [ (x, (aN - x)) | x <- [1 .. aM] ]
            else
                let
                    divN2    = [1 .. (div aN 2) - 1]
                    oddList  = filter (odd) divN2
                    evenList = filter (even) divN2
                    oddPairs =
                        [ (x, (x + y))
                        | (x, y) <- zip [1 .. length oddList] (reverse oddList)
                        ]
                    evenPairs =
                        [ ((aN - x + 1), (aN - x + 1 - y))
                        | (x, y) <- zip [1 .. length evenList]
                                        (reverse evenList)
                        ]
                in
                    take aM (oddPairs ++ evenPairs)



    mapM_ (\x -> putStrLn ((show $fst x) ++ " " ++ (show $snd x))) ret
getVUI :: IO (VU.Vector Int)
getVUI = VU.unfoldr (BSC8.readInt . BSC8.dropWhile isSpace) <$> BSC8.getLine
readI :: BSC8.ByteString -> Int
readI = fst . fromJust . BSC8.readInt
readIL :: BSC8.ByteString -> [Int]
readIL = map readI . BSC8.words
readIT2 :: BSC8.ByteString -> (Int, Int)
readIT2 bs =
    let words = BSC8.words bs in (readI $ head words, readI $ last words)
getI :: IO Int
getI = readI <$> BSC8.getLine
getIL :: IO [Int]
getIL = readIL <$> BSC8.getLine
get2IT :: IO (Int, Int)
get2IT = readIT2 <$> BSC8.getLine

getNIS :: Int -> IO [(Int, BSC8.ByteString)]
getNIS 0 = return []
getNIS n = do
    [aFst, aSnd] <- BSC8.words <$> BSC8.getLine
    next         <- (getNIS (n - 1))
    return ((readI aFst, aSnd) : next)

il2ss :: [Int] -> String
il2ss = unwords . map show

il2sn :: [Int] -> String
il2sn = unlines . map show

modNum = 10 ^ 9 + 7
newtype Mint = M Int
instance Num Mint where
    (M a) + (M b) = M $ mod (a + b) modNum
    (M a) - (M b) = M (a - b)
    (M a) * (M b) = M $ mod (a * b) modNum
    fromInteger a = M $ mod (fromInteger a) modNum
    abs (M a) = M $ abs a
    signum (M a) = M $ signum a
instance Show Mint where
    show (M a) = show (mod a modNum)

int2MInt :: Int -> Mint
int2MInt a = M (mod a modNum)

knappsackMaxVal
    :: AIO.IOArray (Int, Int) Int
    -> VU.Vector Int
    -> VU.Vector Int
    -> Int
    -> IO Int
knappsackMaxVal dp volumeVec valueVec dpSize = do
    mapM_
        (\i -> mapM_
            (\v -> do
                let volume = volumeVec VU.! i
                    value  = valueVec VU.! i
                newdp <- if v < volume
                    then AIO.readArray dp (i, v)
                    else do
                        dpIVm <- AIO.readArray dp (i, v - volume)
                        dpIV  <- AIO.readArray dp (i, v)
                        return (max (dpIVm + value) dpIV)
                AIO.writeArray dp (i + 1, v) newdp
            )
            [0 .. dpSize]
        )
        [0 .. VU.length volumeVec - 1]
    AIO.readArray dp (VU.length volumeVec, dpSize)

knappsackMinVol
    :: AIO.IOArray (Int, Int) Int
    -> VU.Vector Int
    -> VU.Vector Int
    -> Int
    -> Int
    -> IO Int
knappsackMinVol dp volumeVec valueVec dpSize maxVolume = do
    mapM_
        (\i -> mapM_
            (\v -> do
                let volume = volumeVec VU.! i
                    value  = valueVec VU.! i
                newdp <- if (v + value) > dpSize
                    then AIO.readArray dp (i, v)
                    else do
                        dpIVm <- AIO.readArray dp (i, v + value)
                        dpIV  <- AIO.readArray dp (i, v)
                        return (min (dpIVm - volume) dpIV)
                AIO.writeArray dp (i + 1, v) newdp
            )
            (reverse [0 .. dpSize])
        )
        [0 .. VU.length valueVec - 1]

    foldM
        (\acc x -> do
            num <- AIO.readArray dp (VU.length volumeVec, x)
            return $ if num <= maxVolume then x else acc
        )
        0
        [0 .. dpSize]