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: Time Limit Exceeded)

N,M = map(int,input().split())
road = []
near = [[1]]###ゴールへの近道、段階的にリストにしたい。
sign = [0]*(N+1)###簡単のため。最後にsign[0],sign[1]は0になってるから排除できる。
for i in range(M):###入力値の読み取り
    A_i,B_i = map(int, input().split())
    road.append([A_i,B_i])###二重リスト、道の両端がリスト内の値
    #if A_i == 1: ###最後に出力するやつを作ってる。
        #sign[B_i] == 1
       # near[0].append[B_i]
    #if B_i == 1:
       # sign[A_i] == 1
        #near[0].append[A_i]
#print(road)###うまくいってる
j = 0
while len(near[j]) != 0:###near[j]に要素がある時、
    near.append([])
    #print(road)
    for i in near[j]:###near[j]の要素ごとに、要素と繋がってる部屋を検索。色々する
        rem = []###要素ごとにremリストを作る
        #print(i)
        for k in road:###roadでnear[j]すなわち(i)と繋がってる部屋の値を取得(roadの片方がnear[j]の要素なら繋がっている)
            #print(k)
            if k[0] == i:###roadの片方がnear[j]の要素(i)の時
                if sign[k[1]] == 0:###sign[k[1]]が0の時、（上書き防止）
                    sign[k[1]] = i ###roadのもう片方の部屋の道標は(i)
                near[j+1].append(k[1])###nearの次の段階の部屋にroadの(i)じゃない方を追加
                rem.append(road.index(k))###roadの使った要素は後から削除
            if k[1] == i:###roadのもう片方が(i)の時
                if sign[k[0]] == 0:
                    sign[k[0]] = i
                near[j+1].append(k[0])
                rem.append(road.index(k))
        #print(rem)
        #print(road)
        rem.sort(reverse=True)
        for num in rem:###remの数字の要素をroadから削除
            del road[num]
        #print(road)
    j += 1
if len(road) == 0:
    print('Yes')
    for i in range(len(sign)):
        if sign[i] != 0:
            print(sign[i])
else:
    print('No')
#print(road)
#print(near)
#print(sign)###signは上書きされてはいけない