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
open Printf

let ($) g f x = g (f x)

module MyList = struct
  include List
  let foldl = fold_left
  let foldli = fold_lefti
  let foldr = fold_right
  let foldri = fold_righti
  
  let flatmap = concat_map

  let zipf = map2
  let zipfi = map2i
end

module L = MyList

let swap = Tuple2.swap

let atoi = int_of_string
let itoa = string_of_int
let atof = float_of_string
let ftoa = string_of_float
let ctoi = int_of_char
let itoc = char_of_int

let stocl = String.to_list
let cltos = String.of_list
let spsp s = String.split_on_char ' ' s

module MyString = struct
  include String

  let map' f s = L.map f (stocl s)
  let mapi' f s = L.map f (stocl s)

  let zipf f a b = L.zipf f (stocl a) (stocl b)
  let zipfi f a b = L.zipfi f (stocl a) (stocl b)

  let mkstr f sep ls =
    L.foldl (fun acc x -> acc ^ sep ^ (f x)) (f @@ L.hd ls) (List.tl ls)
end

module S = MyString

module MySet (Ord : Set.OrderedType) = struct
  include Set.Make(Ord)

  let fold' f s z = fold f z s
end

module St = MySet

module MyMap (Ord : Map.OrderedType) = struct
  include Map.Make(Ord)

  let apply = find
  let apply_or = find_default
end

module Mp = MyMap


let rds () = read_line ()

let rdi () = atoi @@ rds ()

let rdhs () = spsp @@ rds ()
let rdhi () = rdhs () |> L.map atoi

let rdi2 () = match rdhi () with
  | a::b::_ -> (a, b)
  | _ -> (0, 0)

let rdi3 () = match rdhi () with
| a::b::c::_ -> (a, b, c)
| _ -> (0, 0, 0)

let rdi4 () = match rdhi () with
| a::b::c::d::_ -> (a, b, c, d)
| _ -> (0, 0, 0, 0)

let rdv n rdf =
  let rec rdv' n =
    if n <= 0 then []
    else (
      let s = rdf () in
      s::(rdv' (n - 1))
    )
  in rdv' n

let rdvi n = rdv n rdi
let rdvi2 n = rdv n rdi2
let rdvi3 n = rdv n rdi3
let rdvi4 n = rdv n rdi4

let puts s = print_endline s
let putyn b = puts (if b then "Yes" else "No")
let puti i = puts @@ itoa i

let set_power ls =
  let rec set_power' ls pls =
    match ls with
    | [] -> pls
    | hd::tl -> set_power' tl (L.flatmap (fun p -> [p; hd::p]) pls)
  in set_power' ls [[]]


let rdhf () = rdhs () |> L.map atof
let rdf4 () = match rdhf () with
| a::b::c::d::_ -> (a, b, c, d)
| _ -> (0., 0., 0., 0.)


module Graph = struct
  module Vs = St(Int)
  module Vm = Mp(Int)

  type 'a t = { nb : Vs.t Vm.t; vdt : 'a Vm.t }

  let nothing = { nb = Vm.empty; vdt = Vm.empty }

  let add_edge (e : int * int) g =
    let (v1, v2) = e in
    let nbvs = g.nb |> Vm.apply_or Vs.empty v1 in
    let nb1 = g.nb |> Vm.add v1 (nbvs |> Vs.add v2) in
    { g with nb = nb1 }

  let add_uedge e g =
    add_edge e (add_edge (swap e) g)

  let from_edges es =
    es |> L.foldl (fun g e -> add_edge e g) nothing

  let from_uedges es =
    es |> L.foldl (fun g e -> add_uedge e g) nothing


end

module G = Graph
module Vs = G.Vs
module Vm = G.Vm

let _ =
  let (n, m) = rdi2 () in
  let es = rdv m rdi2 in
  let graph = G.from_uedges es in

  let expand v vp =
    let nb = (graph.nb |> Vm.apply v) in
    nb |> Vs.fold' (fun nbv (nfr, cvp) ->
      if Vm.mem nbv cvp then
        (nfr, cvp)
      else
        (nfr |> Vs.add nbv,
          cvp |> Vm.add nbv v)
    ) (Vs.empty, vp) in
  
  let rec march fr vp =
    if Vs.is_empty fr then
      vp
    else
      let (nfr, nvp) =
        fr |> Vs.fold' (fun v (nfr, cvp) ->
          let (nfr1, nvp) = expand v cvp in
          (Vs.union nfr nfr1, nvp)
        ) (Vs.empty, vp)
      in
      march nfr nvp 
  in

  let rvp = march (Vs.singleton 1) (Vm.singleton 1 0) in
  let result = L.tl @@ Vm.bindings rvp in

  puts "Yes";
  result |> L.iter (fun (v, p)-> puti p)
;;