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)

//#define NDEBUG
#include "bits/stdc++.h"
#include <iostream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <array>
#include <random>
#ifdef _MSC_VER
#include <ppl.h>
//#include <boost/multiprecision/cpp_dec_float.hpp>
//#include <boost/multiprecision/cpp_int.hpp>
//#include <boost/rational.hpp>
//#include <opencv2/core.hpp>
//#include <opencv2/imgproc.hpp>
//#include <opencv2/highgui.hpp>
//#else
//#include <omp.h>
#endif


using namespace std;

#define DUMPOUT cerr
#define dump(...) DUMPOUT<<"  ";DUMPOUT<<#__VA_ARGS__<<" :["<<__LINE__<<":"<<__FUNCTION__<<"]"<<endl;DUMPOUT<<"    ";dump_func(__VA_ARGS__)

using uint = unsigned; using ll = long long; using ull = unsigned long long; using pii = pair<int, int>; using pll = pair<ll, ll>; using pdd = pair<double, double>; using pss = pair<string, string>;
template <typename _KTy, typename _Ty> ostream& operator << (ostream& o, const pair<_KTy, _Ty>& m) { o << "{" << m.first << ", " << m.second << "}"; return o; }
template <typename _KTy, typename _Ty> ostream& operator << (ostream& o, const map<_KTy, _Ty>& m) { if (m.empty()) { o << "{ }"; return o; } o << "{" << *m.begin(); for (auto itr = ++m.begin(); itr != m.end(); itr++) { o << ", " << *itr; } o << "}"; return o; }
template <typename _KTy, typename _Ty> ostream& operator << (ostream& o, const unordered_map<_KTy, _Ty>& m) { if (m.empty()) { o << "{ }"; return o; } o << "{" << *m.begin(); for (auto itr = ++m.begin(); itr != m.end(); itr++) { o << ", " << *itr; } o << "}"; return o; }
template <typename _Ty> ostream& operator << (ostream& o, const vector<_Ty>& v) { if (v.empty()) { o << "{ }"; return o; } o << "{" << v.front(); for (auto itr = ++v.begin(); itr != v.end(); itr++) { o << ", " << *itr; } o << "}"; return o; }
template <typename _Ty> ostream& operator << (ostream& o, const deque<_Ty>& v) { if (v.empty()) { o << "{ }"; return o; } o << "{" << v.front(); for (auto itr = ++v.begin(); itr != v.end(); itr++) { o << ", " << *itr; } o << "}"; return o; }
template <typename _Ty> ostream& operator << (ostream& o, const set<_Ty>& s) { if (s.empty()) { o << "{ }"; return o; } o << "{" << *(s.begin()); for (auto itr = ++s.begin(); itr != s.end(); itr++) { o << ", " << *itr; } o << "}"; return o; }
template <typename _Ty> ostream& operator << (ostream& o, const unordered_set<_Ty>& s) { if (s.empty()) { o << "{ }"; return o; } o << "{" << *(s.begin()); for (auto itr = ++s.begin(); itr != s.end(); itr++) { o << ", " << *itr; }	o << "}"; return o; }
template <typename _Ty> ostream& operator << (ostream& o, const stack<_Ty>& s) { if (s.empty()) { o << "{ }"; return o; } stack<_Ty> t(s); o << "{" << t.top(); t.pop(); while (!t.empty()) { o << ", " << t.top(); t.pop(); } o << "}";	return o; }
template <typename _Ty> ostream& operator << (ostream& o, const list<_Ty>& l) { if (l.empty()) { o << "{ }"; return o; } o << "{" << l.front(); for (auto itr = ++l.begin(); itr != l.end(); ++itr) { o << ", " << *itr; } o << "}"; return o; }
template <typename _KTy, typename _Ty> istream& operator >> (istream& is, pair<_KTy, _Ty>& m) { is >> m.first >> m.second; return is; }
template <typename _Ty> istream& operator >> (istream& is, vector<_Ty>& v) { for (size_t t = 0; t < v.size(); t++) is >> v[t]; return is; }
template <typename _Ty> istream& operator >> (istream& is, deque<_Ty>& v) { for (size_t t = 0; t < v.size(); t++) is >> v[t]; return is; }
namespace aux { // print tuple
  template<typename Ty, unsigned N, unsigned L> struct tp { static void print(ostream& os, const Ty& v) { os << get<N>(v) << ", "; tp<Ty, N + 1, L>::print(os, v); } };
  template<typename Ty, unsigned N> struct tp<Ty, N, N> { static void print(ostream& os, const Ty& v) { os << get<N>(v); } };
}

template<typename... Tys> ostream& operator<<(ostream& os, const tuple<Tys...>& t) { os << "{"; aux::tp<tuple<Tys...>, 0, sizeof...(Tys) - 1>::print(os, t); os << "}"; return os; }

template<typename A, size_t N, typename T> inline void Fill(A(&array)[N], const T& val) { fill((T*)array, (T*)(array + N), val); }

template <typename ... Args>
std::string format(const std::string& fmt, Args ... args) {
  size_t len = std::snprintf(nullptr, 0, fmt.c_str(), args ...);
  std::vector<char> buf(len + 1);
  std::snprintf(&buf[0], len + 1, fmt.c_str(), args ...);
  return std::string(&buf[0], &buf[0] + len);
}

void dump_func() { DUMPOUT << endl; }
template <class Head, class... Tail> void dump_func(Head&& head, Tail&&... tail) { DUMPOUT << head; if (sizeof...(Tail) == 0) { DUMPOUT << " "; } else { DUMPOUT << ", "; } dump_func(move(tail)...); }

#define PI 3.14159265358979323846
#define EPS 1e-8
#define rep(t,n) for(int t=0;t<(n);++t)
#define all(j) (j).begin(), (j).end()
#define SZ(j) ((int)(j).size())
#define fake false

class Timer {
public:
  double t = 0;
  Timer() {}
  static double time() {
#ifdef _MSC_VER
    return __rdtsc() / 2.8e9;
#else
    unsigned long long a, d;
    __asm__ volatile("rdtsc"
      : "=a"(a), "=d"(d));
    return (d << 32 | a) / 2.8e9;
#endif
  }
  void measure() { t = time() - t; }
  double elapsedMs() { return (time() - t) * 1000.0; }
} timer;

struct Xorshift {
  uint64_t x = 88172645463325252LL;
  unsigned next_int() {
    x = x ^ (x << 7);
    return x = x ^ (x >> 9);
  }
  unsigned next_int(unsigned mod) {
    x = x ^ (x << 7);
    x = x ^ (x >> 9);
    return x % mod;
  }
  unsigned next_int(unsigned l, unsigned r) {
    x = x ^ (x << 7);
    x = x ^ (x >> 9);
    return x % (r - l + 1) + l;
  }
  double next_double() {
    return double(next_int()) / UINT_MAX;
  }
} rnd;

template<typename T>
void shuffle_vector(vector<T>& v, Xorshift& rnd) {
  int n = v.size();
  for (int i = n - 1; i >= 1; i--) {
    int r = rnd.next_int(i);
    swap(v[i], v[r]);
  }
}



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

  int N, M, inf = 10000000;
  cin >> N >> M;
  vector<vector<int>> G(N);
  rep(i, M) {
    int a, b;
    cin >> a >> b;
    a--; b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }

  vector<int> d(N, inf), prev(N, -1);
  queue<int> qu;
  d[0] = 0;
  qu.push(0);
  
  while (!qu.empty()) {
    int u = qu.front(); qu.pop();
    for (int v : G[u]) {
      if (d[u] + 1 < d[v]) {
        d[v] = d[u] + 1;
        prev[v] = u;
        qu.push(v);
      }
    }
  }

  cout << "Yes" << endl;
  for (int i = 1; i < N; i++) cout << prev[i] + 1 << endl;

  return 0;
}