Problem: 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.

Examples

Input

4 4
1 2
2 3
3 4
4 2


Output

Yes
1
2
2


Input

6 9
3 4
6 1
2 4
5 3
4 6
1 5
6 2
4 5
5 6


Output

Yes
6
5
5
1
1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

int n, m, ans[100002];
vector<int>adjl[100002];

int main(){
	cin >> n >> m;
	for (int i=0;i<m;i++){
		int u, v;
		cin >> u >> v;
		adjl[u].push_back(v);
		adjl[v].push_back(u);
	}
	memset(ans,-1,sizeof ans);
	queue<pair<int,int>>bfs; // now, prev
	bfs.push({1,0});
	while (!bfs.empty()){
		int now=bfs.front().first;
		int prev=bfs.front().second;
		bfs.pop();
		if (ans[now]!=-1) continue;
		ans[now]=prev;
		for (int next : adjl[now]) bfs.push({next,now});
	}
	cout << "Yes\n";
	for (int i=2;i<=n;i++) cout << ans[i] << "\n";
}