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;
using pr = pair<int, int>;

int N, M, D[100005], pre[100005];
vector<int> v[100005];
queue<int> q;

int main() {
	scanf("%d %d", &N, &M);
	for(int i=0, x, y; i<M; i++) {
		scanf("%d %d", &x, &y);
		v[x].push_back(y), v[y].push_back(x);
	}
	q.push(1);
	for(int i=2; i<=N; i++) D[i] = 1e9;
	while(!q.empty()) {
		int u = q.front(); q.pop();
		for(auto it : v[u]) {
			if(D[it] > D[u] + 1) {
				D[it] = D[u] + 1;
				pre[it] = u;
				q.push(it);
			}
		}
	}
	for(int i=2; i<=N; i++) if(!pre[i]) return puts("No"),0;
	puts("Yes");
	for(int i=2; i<=N; i++) printf("%d\n", pre[i]);
}