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.
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <fstream>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string>
#include <cmath>
#include <cstring>
#include <string>
#include <memory.h>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <ctime>
#include <set>
using namespace std;
vector<int>adj[100005];
vector<vector<int>>ans;
int dfs(int u, int par) {
	int ch=0,nxt=u;
	for(int& v : adj[u]) if(v^par) {
		int endv=dfs(v,u);
		if(endv < 0) continue;
		ch++;
		if(ch==1) nxt=endv;
		else if(ch==2) ans.push_back({par, u, nxt, endv}),nxt=-1;
		else ans.push_back({u,v,v,endv});
	}
	return nxt;
}

int main() {
	int i,to, n;
	for(cin>>n;cin>>n;ans.clear()){
		for(int j=1,x,y;j<n;j++) {
			cin>>x>>y;
			adj[x].push_back(y);
			adj[y].push_back(x);
		}
		for(i=1;;i++) if(adj[i].size()==1) break;
		to=dfs(i,-1);
		cout<<ans.size()<<endl;
		for(auto& x:ans) cout<<x[0]<<" "<<x[1]<<" "<<to<<" "<<x[2]<<endl,to=x[3];
		for(i=1;i<=n;i++) adj[i].clear();
	}
	return 0;
}