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<cstdio>
#include<vector>
using namespace std;
#define N 100005
int n, m, x, y, q[N], c[N], i, j, k;//c[i]=j表示第i号房间指示第j个房号 
vector<int> ac[N];//图的存储 
int main()
{
	scanf("%d%d", &n, &m);
	for(i=1; i<=m; i++){
		scanf("%d%d", &x, &y);
		ac[x].push_back(y);
		ac[y].push_back(x);
	}
	q[1] = 1;
	for(i=j=1; i<=j; i++){
		for(k=0; k<ac[q[i]].size(); k++){
			if(c[ac[q[i]][k]] == 0){//与i相连的第k个房间有没有走过 
				q[++j] =  ac[q[i]][k];
				c[ac[q[i]][k]] = q[i];//入队的肯定是当前从1到目前这个点的最短路径 
			} 
			
		}
	}
	for(i=2; i<=n; i++){
		if(c[i] == 0){
			printf("No");
			return 0;
		} 
	}	
	printf("Yes\n");
	for(i=2; i<=n; i++){
		printf("%d\n", c[i]);
	}
	return 0;
}