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 OCaml (4.10.0) (judge verdict: Accepted)

open Batteries
module EnumL = Enum.Labels
module ListL = List.Labels
module ArrayL = Array.Labels

let dbg0 x = Printf.eprintf "[debug]%s\n" @@ dump x
let dbg1 x = dbg0 x; x

let id = identity

let (++) n m = List.range n `To m
let (++-) n m = if n >= m then List.range n `Downto m else []
let (++^) n m = if n < m then List.range n `To (pred m) else []

let scan fmt = Scanf.sscanf (read_line ()) fmt

let scan_lines n fmt f =
  if n = 0 then []
  else List.map (fun _ -> scan fmt f) (1 ++ n)

let scan_list ?sep cnv =
  let line = read_line () in
  (match sep with
   | None -> List.map String.of_char @@ String.to_list line
   | Some sep -> String.split_on_char sep line)
  |> List.map cnv

let scan_array ?sep cnv = Array.of_list @@ scan_list ?sep cnv

let scan_matrix n m e ?sep conv =
  let arr = Array.make_matrix n m e in
  ListL.iter (0 ++^ n)
    ~f:(fun i -> arr.(i) <- Array.of_list @@ scan_list ?sep conv);
  arr

let atoi c = Char.code c - Char.code '0'

let rec powerset = function
  | [] -> [[]]
  | hd::tl ->
    let pws =  powerset tl in
    pws @ ListL.map pws ~f:(fun pw -> hd::pw)

let permutations l =
  let rec interleave x = function
    | [] -> [[x]]
    | (hd::tl) as l ->
      (x::l) :: (interleave x tl |> ListL.map ~f:(fun l -> hd::l)) in
  let rec aux = function
    | [] -> [[]]
    | a::rest ->
      aux rest |> ListL.map ~f:(interleave a) |> List.concat in
  aux l

let intersection l =
  EnumL.filter ~f:(fun x -> EnumL.exists ~f:((=) x) @@ Enum.clone l)

let zip l m =
  let n = min (List.length l) (List.length m) in
  List.combine (List.take n l) (List.take n m)

let lower_bound n f =
  let rec aux l u =
    if u - l > 1 then
      let m = (l + u) / 2 in
      if f m then aux l m
      else aux m u
    else u in aux (-1) n

let between n x m = n <= x && x < m

let (n, m) = scan "%d %d" Tuple2.make
let ls = scan_lines m "%d %d" Tuple2.make

let () =
  let mat = Array.make (succ n) [] in
  ListL.iter ls ~f:(fun (a,b) ->
      mat.(a) <- b::mat.(a);
      mat.(b) <- a::mat.(b));
  let q = Queue.create () in
  Queue.add (1,1) q;
  let parents = Array.make (succ n) 0 in
  let rec wfs q =
    if Queue.is_empty q then ()
    else
      let (p, n) = Queue.pop q in
      if parents.(n) <> 0 then wfs q
      else
        (parents.(n) <- p;
         ListL.iter mat.(n) ~f:(fun m ->
             Queue.add (n, m) q);
         wfs q)
  in
  wfs q;
  if Array.exists ((=) 0)@@ Array.sub parents 2 (Array.length parents - 2)  then
    Printf.printf "No\n"
  else
    (Printf.printf "Yes\n";
     ListL.iter (2++n) ~f:(fun i ->
         Printf.printf "%d\n" parents.(i)))