Problem:
Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands.
For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands.
Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N.
However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

3 ≤ N ≤ 200 000

1 ≤ M ≤ 200 000

1 ≤ a_i < b_i ≤ N

(a_i, b_i) \neq (1, N)

If i \neq j, (a_i, b_i) \neq (a_j, b_j).

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

POSSIBLE

Sample Input 2

4 3
1 2
2 3
3 4

Sample Output 2

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

100000 1
1 99999

Sample Output 3

IMPOSSIBLE

Sample Input 4

5 5
1 3
4 5
2 3
2 4
1 4

Sample Output 4

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.

Solution in Haskell (GHC 7.10.3) (judge verdict: Time Limit Exceeded)

import qualified Data.Map as Map
import Data.Sequence as Seq
import Control.Monad
import Control.Monad.ST
import Data.STRef

main :: IO ()
main = do
  [n, m] <- (map read . words) <$> getLine :: IO [Int]
  graph <- buildGraph (Map.fromList [] ) m
  let (found, dist) = bfs graph n
  putStrLn $ if found && dist <= 2 then "POSSIBLE" else "IMPOSSIBLE"

buildGraph graph 0 = return graph
buildGraph graph n = do
  [a, b] <- (map read . words) <$> getLine :: IO [Int]
  let graph' = Map.insertWith (++) a [b] graph
  let graph'' = Map.insertWith (++) b [a] graph'
  buildGraph graph'' (n - 1)

bfs :: Map.Map Int [Int] -> Int -> (Bool, Int)
bfs graph n = runST $ do
    visitedRef <- newSTRef (Map.fromList [] :: Map.Map Int Bool)
    found <- newSTRef (False, -1)
    let queue = Seq.fromList [(1, 0)]
    let loop queue =
          case Seq.viewl queue of
            Seq.EmptyL -> return ()
            (a, dist) :< queue' ->
              if dist >= 2
                then return ()
                else do
                  let maybeNeighbors = Map.lookup a graph
                  case maybeNeighbors of
                    Nothing -> undefined
                    Just neighbors ->
                      forM_ neighbors $ \m ->
                        if n == m
                          then do
                            -- ゴールについた
                            writeSTRef found (True, dist + 1)
                            return ()
                          else do
                            visited <- readSTRef visitedRef
                            let v = Map.lookup m visited
                            case v of
                              -- すでに訪れた場所
                              Just True -> return ()
                              -- 新しく訪れた場所
                              Nothing -> do
                                let visited' = Map.insert m True visited
                                writeSTRef visitedRef visited'
                                loop (queue' |> (m, dist + 1))

    loop queue
    readSTRef found