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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<cstdio>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<deque>
#include<tuple>
#include<algorithm>
#include<numeric>
#include<limits>
#include<boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
using namespace std;
#define BEGIN ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define END return EXIT_SUCCESS
#define up(I,A,B) for(auto I=(A);I<=(B);++I)
#define dw(I,A,B) for(auto I=(A);I>=(B);--I)
#define rep(I,N) up(I,0,N-1)
#define all(C) (C).begin(),(C).end()
#define rall(C) (C).rbegin(),(C).rend()
#define ft first
#define sd second
#define mp make_pair
#define mt make_tuple
#define pf push_front
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define rs resize
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template<class T>inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>inline void                                      in(T &p){cin>>p;}
template<class T1,class T2>inline void                            in(T1 &p1,T2 &p2){cin>>p1>>p2;}
template<class T1,class T2,class T3>inline void                   in(T1 &p1,T2 &p2,T3 &p3){cin>>p1>>p2>>p3;}
template<class T1,class T2,class T3,class T4>inline void          in(T1 &p1,T2 &p2,T3 &p3,T4 &p4){cin>>p1>>p2>>p3>>p4;}
template<class T1,class T2,class T3,class T4,class T5>inline void in(T1 &p1,T2 &p2,T3 &p3,T4 &p4,T5 &p5){cin>>p1>>p2>>p3>>p4>>p5;}
template<class T>inline void                                      ins(T &p){for_each(all(p),in<decltype(p[0])>);}
template<class T>inline void                             out(T p){cout<<p<<"\n";}
template<class T1,class T2>inline void                   out(T1 p1,T2 p2){cout<<p1<<" "<<p2<<"\n";}
template<class T1,class T2,class T3>inline void          out(T1 p1,T2 p2,T3 p3){cout<<p1<<" "<<p2<<" "<<p3<<"\n";}
template<class T1,class T2,class T3,class T4>inline void out(T1 p1,T2 p2,T3 p3,T4 p4){cout<<p1<<" "<<p2<<" "<<p3<<" "<<p4<<"\n";}
inline void                                              out(ld p){printf("%.12Lf\n",p);}
template<class T>inline void outml(T p){for_each(all(p),out<decltype(p[0])>);}//ml=multi line
template<class T>inline void outol(T V){auto n=V.size();rep(i,n)cout<<V[i]<<" \n"[i==n-1];}//ol=one line
struct edge {int to,cost;};
typedef pair<int,int> P;
const int MAX_V=1<<18,INF=1<<30;
int V,E,d[MAX_V],r,s,t,c,pre[MAX_V];
vector<edge> G[MAX_V];
priority_queue<P,vector<P>,greater<P>>q;

void dijkstra(int s){
  fill(d,d+V,INF);
  fill(pre,pre+V,-1);
  d[s]=0;
  q.push(P(0,s));
  while(!q.empty()){
    P p=q.top();q.pop();
    int v=p.second;
    if(d[v]<p.first)continue;
    for(int i=0;i<(int)G[v].size();i++){
      edge e=G[v][i];
      if(d[e.to]>d[v]+e.cost){
        d[e.to]=d[v]+e.cost;
        pre[e.to]=v;
        q.push(P(d[e.to],e.to));
      }
    }
  }
}
vector<int> get_path(int t){
  vector<int>path;
  for(;t!=-1;t=pre[t])path.push_back(t);
  reverse(all(path));
  return path;
}
inline void solve(void){
  scanf("%d %d",&V,&E);
  for(int i=0;i<E;i++){
    scanf("%d %d",&s,&t);
    edge e;
    s--;t--;
    if(s<t)swap(s,t);
    e.to=t;e.cost=1;
    G[s].push_back(e);
    e.to=s;
    G[t].push_back(e);
  }
  vector<int>ans;
  up(i,1,V-1){
    dijkstra(i);
    if(d[0]!=INF){
      auto path=get_path(0);
      ans.pb(path[1]);
    }else{
      out("No");
      return;
    }
  }
  out("Yes");
  for(auto e:ans)out(e+1);
}
int main(void){
  BEGIN;
  solve();
  END;
}