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 <bits/stdc++.h>
using namespace std;
#define oo 1000000010
const int N = 200010;

int n;

vector< int > g[N];

vector< pair< int , int > > v;

vector< pair< int , int > > rem;

int dp[N][2];

int last[N] , nxt1[N] , nxt2[N];

pair< int , int > has[N];

void DFS(int node,int prnt){
	last[node] = node;
	int mn1 = oo , mn2 = oo;
	nxt1[node] = nxt2[node] = -1;
	dp[node][0] = dp[node][1] = 0;
	for(int i = 0 ;i < (int)g[node].size();i++){
		if(g[node][i] == prnt) continue;
		DFS(g[node][i], node);
		dp[node][1] += dp[g[node][i]][0] + 1;
		dp[node][0] += dp[g[node][i]][0] + 1;
		if(dp[g[node][i]][1] - dp[g[node][i]][0] - 1 < mn1){
			mn2 = mn1;
			nxt2[node] = nxt1[node];
			mn1 = dp[g[node][i]][1] - dp[g[node][i]][0] - 1;
			nxt1[node] = g[node][i];
		}
		else if(dp[g[node][i]][1] - dp[g[node][i]][0] - 1 < mn2){
			mn2 = dp[g[node][i]][1]  - dp[g[node][i]][0] - 1;
			nxt2[node] = g[node][i];
		}
	}
	if(nxt1[node] == -1){
		has[node] = make_pair(node , node);
	}
	else if(nxt2[node] == -1){
		dp[node][0] += mn1;
		dp[node][1] += mn1;
		last[node] = last[nxt1[node]];
		has[node] = make_pair(node , last[nxt1[node]]);
	}
	else{
		dp[node][0] += mn1 + mn2;
		dp[node][1] += mn1;
		last[node] = last[nxt1[node]];
		has[node] = make_pair(last[nxt1[node]] , last[nxt2[node]]);
	}
}

void buildpath(int node,int prnt,bool start){
	//cout << "HERE " << node << " " << has[node].first << " " << has[node].second << " " << start << endl;
	if(start)
		v.push_back(has[node]);
	for(int i = 0 ;i < (int)g[node].size();i++){
		if(g[node][i] == prnt) continue;
		if(g[node][i] != nxt1[node] && (!start || g[node][i] != nxt2[node])){
			rem.push_back(make_pair(g[node][i] , node));
			buildpath(g[node][i] , node , true);
		}
		else
			buildpath(g[node][i] , node , false);
	}

}


int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i = 1;i <= n;i++) g[i].clear();
		for(int a , b , i = 0 ;i < n - 1;i++){
			scanf("%d%d",&a,&b);
			g[a].push_back(b);
			g[b].push_back(a);
		}
		rem.clear();
		v.clear();
		DFS(1 , -1);
		buildpath(1 , -1 , true);
		printf("%d\n",(int)rem.size());
		for(int i = 0 ;i < (int)rem.size();i++){
			printf("%d %d %d %d\n",rem[i].first , rem[i].second , v[i].second , v[i + 1].first);
		}
	}	
	return 0;
}