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;

#define N 200005

int n, m, i, x, y, pos[N], v[N], ans;
queue<int> Q;
vector<int> a[N];

int main (void)
{
	scanf("%d%d",&n,&m);
	for (; m; m--) {
		scanf("%d%d",&x,&y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	Q.push(1),v[1]=1,pos[1]=-1;
	while (!Q.empty()) {
		x=Q.front(); Q.pop();
		for (i=0; i<a[x].size(); i++) {
			y=a[x][i];
			if (v[y]) continue;
			v[y]=1;
			pos[y]=x;
			Q.push(y);
		}
	}
	ans=-1;
	for (i=1; i<=n; i++) if (!v[i]) {
		ans=1;
		break;
	}
	if (ans==1) puts("No");
	else {
		puts("Yes");
		for (i=2; i<=n; i++) printf("%d\n",pos[i]);
	}
	return 0;
}