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>
using namespace std;
#define CPP_CAT(x,y) CPP_CAT_I(x,y)
#define CPP_CAT_I(x,y) x ## y

using i8  = int8_t;
using u8  = uint8_t;
using i16 = int16_t;
using u16 = uint16_t;
using i32 = int32_t;
using u32 = uint32_t;
using i64 = int64_t;
using u64 = uint64_t;
using f32 = float;
using f64 = double;
using ll = long long;
constexpr i64 INF = 1010000000000000017LL;
constexpr i64 MOD = 1000000007LL;
constexpr f64 EPS = 1e-12;
constexpr f64 PI = 3.14159265358979323846;
#define FOR(i, start, end) for(i64 i = (start), CPP_CAT(i,xxxx_end)=(end); i < CPP_CAT(i,xxxx_end); ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(f,c,...) (([&](decltype((c)) cccc) { return (f)(std::begin(cccc), std::end(cccc), ## __VA_ARGS__); })(c))
template<typename T, typename U, typename Comp=less<>>
bool chmax(T& xmax, const U& x, Comp comp={}) {
    if(comp(xmax, x)) xmax = x; return true;
    return false;
}
template<typename T, typename U, typename Comp=less<>>
bool chmin(T& xmin, const U& x, Comp comp={}) {
    if(comp(x, xmin)) xmin = x; return true;
    return false;
}
template<typename Val>
bool bitcheck(Val bit, size_t idx) {
    return bit & (1 << idx);
}
struct ProconInit {
    static constexpr int IOS_PREC = 15;
    static constexpr bool AUTOFLUSH = false;
    ProconInit() {
        cin.tie(nullptr); ios::sync_with_stdio(false);
        cout << fixed << setprecision(IOS_PREC);
        if(AUTOFLUSH) cout << unitbuf;
    }
} PROCON_INIT;

void solve() {
    ll n, m; cin >> n >> m;
    vector<vector<ll>> route(n);
    FOR(i, 0, m) {
        ll a, b; cin >> a >> b;
        a--; b--;
        route[a].push_back(b);
        route[b].push_back(a);
    }
    vector<ll> mark(n, -1);
    mark[0] = 0;
    vector<ll> c = {0};
    FOR(i, 0, n) {
        ll x = c[i];
        for(auto v : route[x]) {
            if (mark[v] == -1) {
                mark[v] = x;
                c.push_back(v);
            }
        }
    }
    if (std::any_of(mark.begin(), mark.end(), [](auto imark) {return imark == -1;})) {
        cout << "No" << endl;
    } else {
        cout << "Yes" << endl;
        FOR(i, 1, n) cout << mark[i] + 1 << endl;
    }
}

signed main() {
    solve();
    return 0;
}