Problem:
Score: 400 points

Problem Statement

There is a cave.

The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside.

It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage.

Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1.

If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible.

Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 2 \times 10^5

1 \leq A_i, B_i \leq N\ (1 \leq i \leq M)

A_i \neq B_i\ (1 \leq i \leq M)

One can travel between any two rooms by traversing passages.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_M B_M

Output

If there is no way to place signposts satisfying the objective, print No.

Otherwise, print N lines. The first line should contain Yes, and the i-th line (2 \leq i \leq N) should contain the integer representing the room indicated by the signpost in Room i.

Sample Input 1

4 4
1 2
2 3
3 4
4 2

Sample Output 1

Yes
1
2
2

If we place the signposts as described in the sample output, the following happens:

Starting in Room 2, you will reach Room 1 after traversing one passage: (2) \to 1. This is the minimum number of passages possible.

Starting in Room 3, you will reach Room 1 after traversing two passages: (3) \to 2 \to 1. This is the minimum number of passages possible.

Starting in Room 4, you will reach Room 1 after traversing two passages: (4) \to 2 \to 1. This is the minimum number of passages possible.

Thus, the objective is satisfied.

Sample Input 2

6 9
3 4
6 1
2 4
5 3
4 6
1 5
6 2
4 5
5 6

Sample Output 2

Yes
6
5
5
1
1

If there are multiple solutions, any of them will be accepted.

Solution in Haskell (GHC 8.8.3) (judge verdict: Time Limit Exceeded)

{-# OPTIONS_GHC -O2 #-}
{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE LambdaCase #-}
{-# LANGUAGE BangPatterns #-}
module Main where

import           Data.List
import           Control.Monad
import           Debug.Trace
import qualified Data.ByteString.Char8         as BC
import           Data.Maybe                     ( fromJust, fromMaybe )
import qualified Data.Set                      as S
import qualified Data.IntMap.Strict            as IM
import qualified Data.IntMap.Lazy            as IML
import qualified Data.Map.Strict               as M
import           Data.Char
import qualified Data.Array.IArray as A
import qualified Data.Array.Unboxed as UA
import Data.Bits
import Control.Monad.State.Strict
import qualified Data.Array.ST as STA
import qualified Data.Array.IO as IOA
import qualified Data.Array.MArray as MA
import Control.Monad.ST
import System.IO (stdin)
import Control.Arrow
import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as UV
import qualified Data.Vector.Generic as GV
import qualified Data.Vector.Mutable as MV
import Data.Ix
import Data.Ord (Down(Down))
-- import Control.Lens

debug = False

ts :: Show a => a -> a
ts a = if debug then traceShowId a else a


readTuple2 :: IO (Int, Int)
readTuple2 = do
    [a, b] <- readInts
    return (a, b)

readTuple3 :: IO (Int, Int, Int)
readTuple3 = do
    [a, b, c] <- readInts
    return (a, b, c)

readTuple4 :: IO (Int, Int, Int, Int)
readTuple4 = do
    [a, b, c, d] <- readInts
    return (a, b, c, d)

readInt :: IO Int
readInt = fst . fromJust . BC.readInt <$> BC.getLine

readInts :: IO [Int]
readInts = map (fst . fromJust . BC.readInt) . BC.words <$> BC.getLine

split :: (a -> Bool) -> [a] -> [[a]]
split f [] = []
split f as =
    let (fo, la) = span f as
        las      = case la of
            [] -> []
            xs -> split f $ tail xs
    in  case fo of
            [] -> las
            xs -> xs : las

modN = 1000000007

pow :: Num a => a -> Int -> a
pow base 0 = 1
pow base 1 = base
pow base time
    | odd time = let
        a = pow base (div time 2)
        in
        a * a * base
    | otherwise = let
        a = pow base (div time 2)
        in
        a * a

inv :: ModNum -> ModNum
inv a = pow a $ modN - 2

newtype ModNum = ModNum {getInt :: Int} deriving (Show, Eq, Ord)
instance Num ModNum where
    (+) (ModNum a) (ModNum b) = ModNum $ mod (a + b) modN
    (-) (ModNum a) (ModNum b)
            | a < b = (ModNum $ a + modN) - ModNum b
            | otherwise = ModNum $ mod (a - b) modN
    (*) (ModNum a) (ModNum b) = ModNum $ mod (a * b) modN
    abs (ModNum a)
            | a < 0 = abs . ModNum $ a + modN
            | otherwise = ModNum a
    signum (ModNum a) = ModNum $ signum a
    fromInteger i = abs . ModNum . fromIntegral $ mod i (fromIntegral modN)

mDiv :: ModNum -> ModNum -> ModNum
mDiv a b = a * inv b

newtype MyGraph = MyGraph (A.Array Int [Int])
listToGraph :: Int -> [(Int, Int)] -> Bool -> MyGraph
listToGraph n paths inverse = MyGraph $ STA.runSTArray $ do
    a <- MA.newArray (1, n) []
    forM_ paths $ \(from, to) -> do
        f <- MA.readArray a from
        MA.writeArray a from $ to : f
        when inverse $ do
            t <- MA.readArray a to
            MA.writeArray a to $ from : t
    pure a

getPaths :: Int -> MyGraph -> [Int]
getPaths i (MyGraph lst) = lst A.! i

--solve :: Int -> Int -> [(Int, Int)] -> Maybe [Int]
solve n m abList = if -1 `GV.elem` vec then Nothing else Just $ vec  where
  grp = listToGraph n abList True
  go [] vec = pure ()
  go !rooms !vec = do
    !nexts <- filterM (\room -> MV.read vec (room - 1) >>= pure . (== (negate 1)))  (rooms >>= (`getPaths` grp))
    let !nexts' = filter (/=1) $ nub nexts
    forM_ rooms $ \room ->
      forM_ (getPaths room grp) $ \nextRoom -> do
        !val <- MV.read vec (nextRoom - 1)
        when (val == -1) $ MV.write vec (nextRoom - 1) room
    go (ts nexts') vec

  vec = runST $ do
    vec <- MV.replicate n $ negate 1
    go [1] vec
    GV.freeze vec





main :: IO ()
main = do
  (n, m) <- readTuple2
  abList <- replicateM m readTuple2
  case solve n m abList of
    Nothing -> putStrLn "No"
    Just xs -> do
      putStrLn "Yes"
      forM_ [2..n] $ \a -> print $ xs GV.! (a - 1)