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 <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <iomanip>
#include <vector>
#include <string.h>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <math.h>
#include <algorithm>
#include <numeric>


using namespace std;

// マクロ&定数&関数 ================================================
typedef unsigned int uint;
typedef long long ll;

typedef pair<ll, ll> pll;

typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<bool> vbool;
typedef vector<string> vstring;

typedef vector<pair<int, int>> vpint;
typedef vector<pair<ll, ll>> vpll;
typedef vector<pair<double, double>> vpdouble;

typedef vector<vector<int>> vvint;
typedef vector<vector<ll>> vvll;
typedef vector<vpint> vvpint;
typedef vector<vpll> vvpll;
typedef vector<vector<double>> vvdouble;
typedef vector<vector<string>> vvstring;
typedef vector<vector<bool>> vvbool;

typedef vector<map<ll, ll>> vmll;

typedef vector<vector<vector<ll>>> vvvll;

const int INF = 1e9 + 1;
const ll LLINF = 1e17 + 1;
const int DX[9] = { 0,  0, 1, -1, 1,  1, -1, -1, 0 }; // 4;４近傍
const int DY[9] = { 1, -1, 0,  0, 1, -1,  1, -1, 0 }; // 8:８近傍 9:(0,0)を含む
const ll MOD = 1e9 + 7; //10^9 + 7
const ll MAX = 1e9;
const double pllI = 3.14159265358979323846264338327950288;



//---------------------------------------------------------------
//  オーバーフローチェック
//---------------------------------------------------------------
bool is_overflow(ll a, ll b)
{
	return ((a * b) / b != a);
}

//---------------------------------------------------------------
//  約数列挙
//---------------------------------------------------------------
vll divisor(ll n) {
	vll ret;
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(begin(ret), end(ret));
	return (ret);
}

//---------------------------------------------------------------
// N以下のすべての素数を列挙する(エラトステネスの篩)
//---------------------------------------------------------------
vbool searchSosuu(ll N) {
	vbool sosuu; // ←これをグローバル変数にして１度だけ実行する
	for (ll i = 0; i < N; i++) {
		sosuu.emplace_back(true);
	}
	sosuu[0] = false;
	sosuu[1] = false;
	for (ll i = 2; i < N; i++) {
		if (sosuu[i]) {
			for (ll j = 2; i * j < N; j++) {
				sosuu[i * j] = false;
			}
		}
	}

	return sosuu;
}

//---------------------------------------------------------------
// 素因数分解　O(√N)
//---------------------------------------------------------------
vpll div_prime(ll n)
{
	vpll prime_factor;
	for (ll i = 2; i * i <= n; i++)
	{
		ll count = 0;
		while (n % i == 0)
		{
			count++;
			n /= i;
		}
		if (count)
		{
			pair<ll, ll> temp = { i, count };
			prime_factor.emplace_back(temp);
		}
	}
	if (n != 1)
	{
		pair<ll, ll> temp = { n, 1 };
		prime_factor.emplace_back(temp);
	}

	return prime_factor;
}

//---------------------------------------------------------------
// 素数判定
//---------------------------------------------------------------
bool is_sosuu(ll N) {

	if (N < 2) {
		return false;
	}
	else if (N == 2) {
		return true;
	}
	else if (N % 2 == 0) {
		return false;
	}

	for (ll i = 3; i <= sqrt(N); i += 2) {
		if (N % i == 0) {
			return false;
		}
	}

	return true;
}

//---------------------------------------------------------------
// 最大公約数（ユークリッドの互除法）(O(logA))
//---------------------------------------------------------------
ll gcd(ll a, ll b) {
	if (a < b) {
		ll tmp = a;
		a = b;
		b = tmp;
	}

	ll r = a % b;
	while (r != 0) {
		a = b;
		b = r;
		r = a % b;
	}

	return b;
}

//---------------------------------------------------------------
// 最小公倍数
//---------------------------------------------------------------
ll lcm(ll a, ll b) {
	ll temp = gcd(a, b);
	return temp * (a / temp) * (b / temp);
}

//---------------------------------------------------------------
// 階乗
//---------------------------------------------------------------
ll factorial(ll n)
{
	if (n <= 1)
	{
		return 1;
	}

	return (n * (factorial(n - 1))) % MOD;
}

//---------------------------------------------------------------
// 高速コンビネーション計算(前処理:O(N) 計算:O(1))
//---------------------------------------------------------------
// テーブルを作る前処理
ll comb_const = 200005;
vll fac(comb_const), finv(comb_const), inv(comb_const);
bool COMineted = false;
void COMinit() {

	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (ll i = 2; i < comb_const; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}

	COMineted = true;
}

// 二項係数計算
ll COM(ll n, ll k) {
	if (COMineted == false) COMinit();

	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

//---------------------------------------------------------------
// 繰り返し２乗法 base^sisuu
//---------------------------------------------------------------
ll RepeatSquaring(ll base, ll sisuu) {
	if (sisuu < 0) {
		cout << "RepeatSquaring: 指数が負です!" << endl;
		return 0;
	}
	if (sisuu == 0) return 1;
	if (sisuu % 2 == 0) {
		ll t = RepeatSquaring(base, sisuu / 2) % MOD;
		return (t * t) % MOD;
	}
	return base * RepeatSquaring(base, sisuu - 1) % MOD;
}

//---------------------------------------------------------------
// 高速単発コンビネーション計算(O(logN))
//---------------------------------------------------------------
ll fast_com(ll a, ll b) {

	ll bunshi = 1;
	ll bunbo = 1;
	for (ll i = 1; i <= b; i++)
	{
		bunbo *= i;
		bunbo %= MOD;

		bunshi *= (a - i + 1);
		bunshi %= MOD;
	}

	ll ret = bunshi * RepeatSquaring(bunbo, MOD - 2);

	ret %= MOD;
	while (ret < 0)
	{
		ret += MOD;
	}

	return ret;
}

//---------------------------------------------------------------
// 2直線の交差判定(直線(x1, y1)->(x2, y2) と 直線(X1, Y1)->(X2, Y2))
//---------------------------------------------------------------
bool is_cross(ll x1, ll y1, ll x2, ll y2, ll X1, ll Y1, ll X2, ll Y2)
{
	ll dx_ai = X1 - x1;
	ll dy_ai = Y1 - y1;
	ll dx_bi = X1 - x2;
	ll dy_bi = Y1 - y2;

	ll dx_ai2 = X2 - x1;
	ll dy_ai2 = Y2 - y1;
	ll dx_bi2 = X2 - x2;
	ll dy_bi2 = Y2 - y2;

	ll si = dx_ai * dy_bi - dy_ai * dx_bi;
	ll si2 = dx_ai2 * dy_bi2 - dy_ai2 * dx_bi2;

	ll si3 = dx_ai * dy_ai2 - dy_ai * dx_ai2;
	ll si4 = dx_bi * dy_bi2 - dy_bi * dx_bi2;

	return (si * si2 < 0 && si3 * si4 < 0);

}

//---------------------------------------------------------------
// 最長増加部分列の長さ(O(NlogN))
//---------------------------------------------------------------
ll LSI(vll vec, ll size)
{
	vll lsi(size + 1); // 長さjを作った時の右端の最小値
	for (ll i = 0; i <= size; i++)
	{
		lsi[i] = LLINF;
	}

	lsi[0] = 0;
	lsi[1] = vec[0];

	for (ll i = 1; i < size; i++)
	{
		// 初めてvec[i]の方が小さくなるところを探す
		auto Iter = lower_bound(lsi.begin(), lsi.end(), vec[i]);
		ll idx = Iter - lsi.begin();

		if (idx > 0 && lsi[idx - 1] < vec[i]) // 条件文の前半怪しい
		{
			lsi[idx] = vec[i];
		}
	}

	for (ll i = size; i >= 0; i--)
	{
		if (lsi[i] < LLINF)
		{
			return i;
		}
	}
}

//---------------------------------------------------------------
// 木の根からの深さ
//---------------------------------------------------------------
vll tree_depth(vvll edge, ll start_node, ll n_node)
{
	vll dist(n_node, LLINF);
	dist[start_node] = 0;

	stack<pll> S;
	S.push({ start_node, 0 });

	while (!S.empty())
	{
		ll node = S.top().first;
		ll d = S.top().second;
		dist[node] = d;

		S.pop();

		for (int i = 0; i < edge[node].size(); i++)
		{
			if (dist[edge[node][i]] == LLINF)
			{
				S.push({ edge[node][i], d + 1 });
			}
		}
	}

	return dist;
}


//---------------------------------------------------------------
// ワーシャルフロイド法(O(N^3)) 任意の2点間の最短距離
//---------------------------------------------------------------
vvll warshall_floyd(ll n, vvll d) {
	for (int k = 0; k < n; k++) {       // 経由する頂点
		for (int i = 0; i < n; i++) {    // 始点
			for (int j = 0; j < n; j++) {  // 終点
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
			}
		}
	}
	return d;
}

//---------------------------------------------------------------
// Union Find
//---------------------------------------------------------------
/*
	UnionFind uf(要素の個数);
	for(int i = 0;i < 関係の個数; i++)
	{
		uf.merge(A[i], B[i]);
	}

	nを含む集合の大きさ = uf.size(n)
	nを含む集合の代表者 = uf.root(n)
	集合の個数 = uf.n_group
*/
class UnionFind {
public:
	vector <ll> par; // 各元の親を表す配列
	vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)
	ll n_group; //集合の数

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
		n_group = sz_;
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		n_group--;
		return true;
	}

	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}
};


//========================================================================


struct edge {
	int to, cost;
};
typedef vector<vector<edge> > AdjList;
AdjList graph;

typedef pair<int, int> P;


vector<int> dist;
vector<int> prever;

void dijkstra(int n, int s) {
	dist = vector<int>(n, INF);
	prever = vector<int>(n, -1);
	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();
		int v = p.second;
		if (dist[v] < p.first) {
			continue;
		}
		for (int i = 0; i < graph[v].size(); i++) {
			edge e = graph[v][i];
			if (dist[e.to] > dist[v] + e.cost) {
				dist[e.to] = dist[v] + e.cost;
				prever[e.to] = v;
				que.push(P(dist[e.to], e.to));
			}
		}
	}
}

vector<int> get_path(int t) { //頂点tへの最短路
	vector<int> path;
	for (; t != -1; t = prever[t]) {
		path.push_back(t);
	}

	reverse(path.begin(), path.end());
	return path;
}

int main(void) {
	int n, m;
	cin >> n >> m;
	graph = AdjList(n);
	for (int i = 0; i < m; i++) {
		int from, to, cost;
		cin >> from >> to;
		from--; to--;
		edge t = { to, 1 };
		edge t2 = { from, 1 };
		graph[from].push_back(t);
		graph[to].push_back(t2);
	}

	dijkstra(n, 0);

	vll ans(n, -1);
	ans[0] = -1;

	vbool ok(n, false);
	ok[0] = true;

	for (int i = 1; i < n; i++)
	{
		if (!ok[i])
		{
			vint rout = get_path(i);
			reverse(rout.begin(), rout.end());

			for (int j = 0; j < rout.size() - 1; j++)
			{
				if (ans[rout[j]] == -1)
				{
					ans[rout[j]] = rout[j + 1];
				}
				else
				{
					break;
				}
			}
		}
	}

	for (int i = 1; i < n; i++)
	{
		if (ans[i] == -1)
		{
			cout << "No";
			return 0;
		}
	}

	cout << "Yes" << endl;
	for (int i = 1; i < n; i++)
	{
		cout << ans[i] + 1 << endl;
	}



}