Problem: 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.

Examples

Input

4 4
1 2
2 3
3 4
4 2


Output

Yes
1
2
2


Input

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


Output

Yes
6
5
5
1
1

Solution (in CPP):

Solution:
/*

*/

#include<bits/stdc++.h>
using namespace std;
typedef pair<long,long> P;
int main(){
  long N,M;cin>>N>>M;
  vector<vector<long>>A(N);
  for(long i=0;i<M;i++){
    long a,b;cin>>a>>b;
    A[a-1].push_back(b-1);
    A[b-1].push_back(a-1);
  }
  vector<long>D(N,-1);D[0]=0;
  queue<long>q;q.push(0);
  while(q.size()){
    long f=q.front();q.pop();
    for(long x:A[f])if(D[x]==-1){
      D[x]=f;
      q.push(x);
    }
  }
  for(long x:D)if(x==-1){cout<<"No";return 0;}
  cout<<"Yes"<<endl;
  for(long i=1;i<N;i++)cout<<D[i]+1<<endl;
  
  
}