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: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep1(i, n) for (int i = 1; i <= (n); i++)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<vector<ll>> Matrix;

const char wall = '#';
vector<vector<char>> fields;
vector<vector<bool>> used;
vector<vector<ll>> dist;
const int dx[4] = {0, -1, 0, 1};
const int dy[4] = {1, 0, -1, 0};

vector<ll> A;
// vectorのリサイズ
// A.resize(N);

ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a*b/gcd(a, b);}


struct edge {
  long long to;
  long long cost;
};

const ll INF = 100000000;
struct Graph {
  ll V;
  vector<vector<edge>> G;
  vector<ll> dist;
  // prev[to] = from
  // to に到達前にfromにいたよ
  vector<ll> prev;
 
  // construct
  //
  // usage: graph g(n);
  Graph(ll n) {
    init(n);
  }

  void init(ll n) {
    V = n;
    G.resize(V);
    dist.resize(V, INF);
    prev.resize(V, -1);
  }

  // 頂点sから頂点tへコストcostの辺が張られる
  void add_edge(ll s, ll t, ll cost) {
    edge e;
    e.to = t; e.cost = cost;
    G[s].push_back(e);
  }

  // 頂点sから各頂点への最短経路をダイクストラ法で探索
  // debug = true でデバッグ出力する
  //
  // コストが負の辺があると使えないから注意
  //
  // 頂点iへの距離は ll dist = d.dist[i]
  //
  // O(E + VlogV)
  void dijkstra(ll s, bool debug) {
    dist[s] = 0;
    priority_queue<P, vector<P>, greater<P>> que;
    que.push(P(0, s));
    while(!que.empty()) {
      P p = que.top(); que.pop();
      ll v = p.second;
      if (debug) printf("v: %lld, dist[v]: %lld, cost: %lld\n", v, dist[v], p.first);
      if (dist[v]<p.first) {
        if (debug) printf("v: %lld, dist[v]: %lld, p.first: %lld skip", v, dist[v], p.first);
        continue;
      }
      for (auto nextE : G[v]) {
        if (dist[nextE.to] > dist[v] + nextE.cost) {
          if (debug) printf("v: %lld, next: %lld\n", v, nextE.to);
          dist[nextE.to] = dist[v] + nextE.cost;
          prev[nextE.to] = v;
          que.push(P(dist[nextE.to], nextE.to));
        }
      }
    }
  }

  // 頂点sから各頂点への最短経路をベルマンフォード法で探索
  //
  // 戻り値がtrueならば負の閉路を含む
  bool bellman_ford(int s) {
    dist[s] = 0;
    rep(i, V) {
      rep(v, V) {
        rep(k, G[v].size()) {
          edge e = G[v][k];
          if (dist[v] != INF && dist[e.to] > dist[v] + e.cost) {
            dist[e.to] = dist[v] + e.cost;
            if ( i == V - 1 ) return true;
          }
        }
      }
    }
    return false;
  }

  // 頂点tまでの経路を返す
  // ダイクストラ法で探索した場合のみ使える
  vector<ll> get_path(ll t) {
    vector<ll> path;
    for (ll cur = t; cur != -1; cur = prev[cur]) {
      path.push_back(cur);
    }
    reverse(path.begin(), path.end());
    return path;
  }

};


int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  ll n, m;
  cin >> n >> m;
  Graph g(n);
  rep(i, m) {
    ll a, b;
    cin >> a >> b;
    a--; b--;
    g.add_edge(a, b, 1);
    g.add_edge(b, a, 1);
  }

  //rep(i, n) {
  //  rep(j, g.G[i].size()) {
  //    printf("%lld%c", g.G[i][j].to, (i ==  g.G[i].size()-1)?'\n':' ');
  //  }
  //}
  g.dijkstra(0, false);
  rep(i, n) {
  	// unreachable のときは INFが入っている
  // unreachableの考慮を忘れないこと
    if (g.dist[i] == INF) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  for (ll i = 1; i < n; i++ ) {
    cout << g.prev[i]+1 << endl;
  }

  //rep(i, n) {
  //  cout << i << ", " << g.dist[i] << ", "  << g.prev[i]<<  endl;
  //}
  // // ダイクストラのときは最短経路わかる
  //vector<ll> path = g.get_path(3);
  //rep(i, path.size()) cout << path[i] << endl;
  return 0;
}