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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.math.BigInteger;
import java.util.*;
import java.io.*;
import java.text.*;

public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 32768);
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer t;

    static String sn() {
        while (t == null || !t.hasMoreTokens()) {
            try {
                t = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }
        return t.nextToken();
    }

    static int ni() {
        return Integer.parseInt(sn());
    }

    static long nlo() {
        return Long.parseLong(sn());
    }

    public static void main(String args[]) {
        int t = 1;
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    static long mod = (long) 998244353;

    static long pro(long a, long b) {
        return (a % mod * b % mod) % mod;
    }

    static long add(long a, long b) {
        return (a + b) % mod;
    }

    static long sub(long a, long b) {
        return ((a % mod - b % mod) + mod) % mod;
    }

    static int maxn = 1000005;

    static int pow(int n) {
        return (int) (Math.log10(n) / Math.log10(2));
    }

    static long modpow(long x, long y) {
        long res = 1l;
        x %= mod;
        while (y > 0) {
            if (y % 2 != 0)
                res = pro(res, x);
            x = pro(x, x);
            y /= 2;
        }
        return res;
    }

    static class pair {
        String a;
        int b;

        pair(String x, int y) {
            a = x;
            b = y;
        }
    }

    static long modInverse(long n) {
        BigInteger n1 = new BigInteger(Long.toString(n));
        String s = (n1.modInverse(new BigInteger(Long.toString(mod)))).toString();
        return Long.parseLong(s);
    }

    static boolean v[] = new boolean[1000005];

    static void seive() {
        v[1] = true;
        for (int i = 2; i < 1000005; i++) {
            if (!v[i]) {
                for (int j = 2 * i; j < 1000005; j += i)
                    v[j] = true;
            }
        }
    }

    static boolean isprime(long n) {
        if (n == 1)
            return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static DecimalFormat df = new DecimalFormat("0.000000000000000000000");

    static void solve() {
        int n = ni();
        int m = ni();
        int p[] = new int[n];
        List<Integer> list[] = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            p[i] = -1;
            list[i] = new ArrayList<>();
        }
        for (int i = 0; i < m; i++) {
            int u = ni() - 1;
            int v = ni() - 1;
            list[u].add(v);
            list[v].add(u);
        }
        Queue<Integer> q = new LinkedList<>();
        q.add(0);
        boolean v[] = new boolean[n];
        v[0] = true;
        while (!q.isEmpty()) {
            int t = q.poll();
            for (int ver : list[t]) {
                if (!v[ver]) {
                    p[ver] = t;
                    q.add(ver);
                    v[ver] = true;
                }
            }
        }
//        PriorityQueue<Integer> pq = new PriorityQueue<>((x, y) -> dp[x] - dp[y]);
//        pq.add(0);
//        while (!pq.isEmpty()) {
//            int t = pq.poll();
//            v[t] = true;
//            for (int h : list[t]) {
//                if (!v[h]) {
//                    if (dp[h] > dp[t] + 1) {
//                        dp[h] = dp[t] + 1;
//                        p[h] = t;
//                    }
//                    pq.add(h);
//                }
//            }
//        }
//        for (int i = 0; i < n; i++) {
//            if (dp[i] == Integer.MAX_VALUE) {
//                out.println("No");
//                return;
//            }
//        }
        for (int i = 1; i < n; i++) {
            if (!v[i]) {
                out.println("No");
                return;
            }
        }
        out.println("Yes");
        for (int i = 1; i < n; i++)
            out.println(p[i] + 1);
    }
}