Swiper, no swiping!
Contest: Codeforces Round 594 (Div. 1)
Difficulty rating: 3400
Tags: graphs, implementation

In anticipation of new adventures Boots wanted to do a good deed. After discussion with the Map and Backpack, they decided to gift Dora a connected graph. After a long search, Boots chose $$$t$$$ graph's variants, which Dora might like. However fox Swiper wants to spoil his plan.

The Swiper knows, that Dora now is only able to count up to $$$3$$$, so he has came up with a following idea. He wants to steal some non-empty set of vertices, so that the Dora won't notice the loss. He has decided to steal some non-empty set of vertices, so that after deletion of the stolen vertices and edges adjacent to them, every remaining vertex wouldn't change it's degree modulo $$$3$$$. The degree of a vertex is the number of edges it is adjacent to. It would've been suspicious to steal all the vertices, so Swiper needs another plan.

Boots are sure, that the crime can not be allowed. However they are afraid, that they won't be able to handle this alone. So Boots decided to ask for your help. Please determine for every graph's variant whether the Swiper can perform the theft or not.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100\,000$$$) — the number of graph variants.

The first line of each variant contains integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 500\,000$$$, $$$0 \le m \le 500\,000$$$), the number of vertexes and edges in the graph.

Then $$$m$$$ lines follow, each containing integers $$$a_i$$$, $$$b_i$$$ ($$$1 \le a_i, b_i \le n$$$), the indices of the vertices connected with a corresponding edge.

It's guaranteed, that the graph is connected and doesn't contain multiple edges or self-loops.

It's guaranteed, that the sum of $$$n$$$ over all variants is at most $$$500\,000$$$ and that the sum of $$$m$$$ over all variants is at most $$$500\,000$$$.

Descriptions of graph's variants are separated with an empty line.

Output format:
For each variant:

- In case the answer exists, print "Yes" and then the answer itself.The first line should contain an integer $$$c$$$ ($$$1 < c < n$$$), the number of vertices the Crook can steal, without Dora noticing the loss. On the next line print $$$c$$$ distinct integers, the indices of the graph's vertices in arbitrary order.
- Otherwise print "No".

In case there are several correct ways to steal the vertices, print any of them.

Please note, that it's not required to maximize the number of stolen vertices.

Constraints:
Time limit: 2.0 s
Memory limit: 512.0 MB

Note:
The picture below shows the third variant from the example test. The set of the vertices the Crook can steal is denoted with bold.

Example input:
3
3 3
1 2
2 3
3 1

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

8 12
1 2
1 3
2 3
1 4
4 5
5 1
3 6
3 7
3 8
6 1
7 1
8 1

Example output:
No
Yes
3
4 5 6
Yes
3
6 7 8

Editorial:
We will look only on set of vertex, that will stay after deleting.
Let node-$$$x$$$ — node with degree $$$mod\ 3$$$ equal $$$x$$$.
Let's consider some cases:
• Node-$$$0$$$ exists. It is the answer, except the case, when our graph consists of one node.
• Cycle on nodes-$$$2$$$ exists. So exists irreducible cycle on nodes-$$$2$$$. It is the answer, except the case, when our graph is cycle.
• Way between nodes-$$$1$$$ on nodes-$$$2$$$ exists. So exists irreducible way with same conditions. It is the answer, except the case, when our graph is way.
• Our graph contains one node-$$$1$$$ and nodes-$$$2$$$ form forest. Let's take two trees and delete all except two ways in this trees, such only endpoints are connected with node-$$$1$$$. Those ways and node-$$$1$$$ are the answer, except the case, when our graph is twho cycles with one common node.
That's all cases. Let's show it:
• Nodes-$$$2$$$ form forest. Exists at least one leaf, that is connected with node-$$$1$$$. So in that case node-$$$1$$$ always exists.
• Degree of node-$$$1$$$ equals sum degree of nodes-$$$2$$$ minus doubled number of edges between nodes-$$$2$$$. Let $$$k$$$ — number of nodes-$$$2$$$. So degree of node-$$$1$$$ $$$mod\ 3$$$ equals $$$2k - 2(k - 1) = 2$$$. Contradiction. So forest consist of at least two trees.