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 Python (3.8.2) (judge verdict: Accepted)

import queue
N,M = map(int,input().split())
adj = [[]for i in range(N+1)] #どこの部屋とどこの部屋が繋がっているかを表す、N＋1の理由は、部屋番号とインデックスが一個ずれてるから
parent = [-1] * (N+1) #-1なら未訪問、N+1な理由は同様である
for m in range(M):
    a,b = map(int,input().split())
    adj[a].append(b) #お互いの場所を書く、繋がっているかを記録する
    adj[b].append(a) #こっちにもお互いの場所を書く
q = queue.Queue() #記録するようのトンネルみたいなもの、入ったものから出ていく感じ、自分の部屋番号、親の部屋番号（ただ作っただけ）
q.put((1,0)) # 部屋1の親は部屋0と考える（二個ぶち込んでる) 
while(not q.empty()): #もう探索の候補がqueueに入ってこなくなったら終わり
    x, p = q.get() #探索の候補を受け取る、一回取り出したらもう抜ける
    if parent[x] != -1: #-1じゃなくて変えられていたら訪問済みでスキップ
        continue
    parent[x] = p #親を示しているリストを更新する
    for next in adj[x]: #今いる場所に繋がっている部屋を全てループする
        if parent[next] == -1: # 未訪問なら
            q.put((next,x)) # 探索候補に追加、これがもう全て-1じゃなくて回りきったら終わりとする
print ('Yes') # 必ず目標は達成できる
for p in parent[2:]: # 部屋2以降の親（前の部屋）
    print(p)