Problem: Nastia has an unweighted tree with n vertices and wants to play with it!

The girl will perform the following operation with her tree, as long as she needs:

  1. Remove any existing edge. 
  2. Add an edge between any pair of vertices. 



What is the minimum number of operations Nastia needs to get a bamboo from a tree? A bamboo is a tree in which no node has a degree greater than 2.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5) — the number of vertices in the tree.

Next n - 1 lines of each test cases describe the edges of the tree in form a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i).

It's guaranteed the given graph is a tree and the sum of n in one test doesn't exceed 2 ⋅ 10^5.

Output

For each test case in the first line print a single integer k — the minimum number of operations required to obtain a bamboo from the initial tree.

In the next k lines print 4 integers x_1, y_1, x_2, y_2 (1 ≤ x_1, y_1, x_2, y_{2} ≤ n, x_1 ≠ y_1, x_2 ≠ y_2) — this way you remove the edge (x_1, y_1) and add an undirected edge (x_2, y_2).

Note that the edge (x_1, y_1) must be present in the graph at the moment of removing.

Example

Input


2
7
1 2
1 3
2 4
2 5
3 6
3 7
4
1 2
1 3
3 4


Output


2
2 5 6 7
3 6 4 5
0

Note

Note the graph can be unconnected after a certain operation.

Consider the first test case of the example: 

<image> The red edges are removed, and the green ones are added.

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int max_n = 1e5+100;
vector<int> G[max_n];
vector<int> v1,v2,v3,v4;
int n;

int dfs(int u,int p)
{
	int du=0;int net = u;
	for (int v:G[u])
	{
		if (v==p)continue;
		int re = dfs(v,u);
		if (re<0)continue;
		++du;
		if (du==1)net = re;
		else if (du==2)
		{
			v1.push_back(p);
			v2.push_back(u);
			v3.push_back(net);
			v4.push_back(re);
			net=-1;
		}
		else
		{
			v1.push_back(u);
			v2.push_back(v);
			v3.push_back(v);
			v4.push_back(re);
		}
	}
	return net;
}

int main()
{
	ios::sync_with_stdio(0);
	int t;cin>>t;
	while (t--)
	{
		cin>>n;
		for (int i=0;i<=n;++i)G[i].clear();
		v1.clear();v2.clear();v3.clear();v4.clear();
		for (int i=1,u,v;i<n;++i)
		{
			cin>>u>>v;
			G[u].push_back(v);
			G[v].push_back(u);
		}
		int to;
		for (int i=1;i<=n;++i)
		{
			if (G[i].size()==1)
			{
				to=dfs(i,-1);
				break;
			}
		}
		cout<<v1.size()<<endl;
		for (int i=0;i<v1.size();++i)
		{
			cout<<v1[i]<<" "<<v2[i]<<" "<<to<<" "<<v4[i]<<endl;
			to=v3[i];
		}
	}
}