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>
#define pii pair<int,int>
#define fi first
#define se second
#define re register
using namespace std;
typedef long long ll;

int rd(){
	int x = 0;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x;
}

vector<pair<pii,pii>> choose;
const int maxn = 1e5 + 5;
vector<pii> e[maxn];
int siz[maxn],vis[maxn];
int del[maxn];
int ans;

void dfs(int u,int fa){
	int cnt = 0;
	siz[u] = 1;
	for(auto v:e[u]) {
		if(v.fi == fa) continue;
		dfs(v.fi,u);
		siz[u] += siz[v.fi];
		if(!vis[v.fi]) cnt++;
		else {
			ans++;
			del[v.se] = 1;
		}
	}
	if(cnt >= 2) {
		siz[u] -= 2;
		vis[u] = 1;
		for(auto v:e[u]) {
			if(v.fi == fa) continue;
			if(cnt <= 2) break;
			if(!vis[v.fi]) {
				cnt--;
				ans++;
				del[v.se] = 1;
			}
		}
	}
	else if(cnt == 1) siz[u]--; 
}

vector<pii> bamboos;
vector<int> leaves;
int used[maxn];

void dfs2(int u,int rt){
	used[u] = 1;
	int cnt = 0;
	for(auto v:e[u]) {
		if(used[v.fi] || del[v.se]) continue;
		cnt++;
		dfs2(v.fi,rt);
	}
	if(u == rt && cnt == 1) 
	   leaves.push_back(u);
	else if(!cnt) 
		leaves.push_back(u);	
}


int main(){
	int T = rd();
	while(T--){
		ans = 0;
		int n = rd();
		for(int i = 0;i <= n;i++) e[i].clear(),used[i] = vis[i] = siz[i] = del[i] = 0;
		leaves.clear();
		bamboos.clear();
		choose.clear();
		for(int i = 1;i < n;i++){
			int x = rd();
			int y = rd();
			e[x].push_back(make_pair(y,i));
			e[y].push_back(make_pair(x,i));
		}
		dfs(1,1);
		for(int i = 1;i <= n;i++){
			if(!used[i]) {
				dfs2(i,i);
				if(leaves.size() == 2) {
					bamboos.push_back(make_pair(leaves[0],leaves[1]));
				}
				else bamboos.push_back(make_pair(leaves[0],leaves[0]));
				leaves.clear();
			}
		}
		vector<pii> Del,Add;
		for(int i = 1;i <= n;i++){
			for(auto v:e[i]) {
				if(del[v.se]) {
					if(i < v.fi)
						Del.push_back(make_pair(i,v.fi));
				}
			}	
		}
		for(int i = 1;i < bamboos.size();i++){
			Add.push_back(make_pair(bamboos[i - 1].se,bamboos[i].fi));
		}
		for(int i = 0;i < ans;i++)
			choose.push_back(make_pair(Del[i],Add[i]));
		cout << ans << '\n';
		for(auto it:choose){
			cout << it.fi.fi << ' ' << it.fi.se << ' ' << it.se.fi << ' ' << it.se.se << '\n'; 
		}
	}
}