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<set>
#include<cstdio>
#include<iostream>
#include<algorithm>

using namespace std;

#define N 200001

int n;
int front[N],to[N<<1],nxt[N<<1],tot;
int front2[N],to2[N<<1],nxt2[N<<1],tot2;

int cut[N][2],ans;
int con[N][2];

bool vis[N];
int en;

int d[N];

set<long long>edge;
const long long mod=1e6;

void add(int u,int v)
{
	to[++tot]=v; nxt[tot]=front[u]; front[u]=tot;
	d[u]++;
}

void add2(int u,int v)
{
	to2[++tot2]=v; nxt2[tot2]=front2[u]; front2[u]=tot2;
}

void cutdown(int x,int y)
{
	if(x>y) swap(x,y);
	long long m=x*mod+y;
	edge.insert(m);
	ans++;
	cut[ans][0]=x;
	cut[ans][1]=y;
	d[x]--;
	d[y]--;
}

bool search(int x,int y)
{
	if(x>y) swap(x,y);
	long long m=x*mod+y;
	if(edge.find(m)!=edge.end()) return true;
	return false;
}

void dfs(int x,int fa,bool havefa)
{
	int ch=0,t;
	for(int i=front[x];i;i=nxt[i]) 
	{
		t=to[i];
		if(t==fa) continue;
		dfs(t,x,1);
	}
	for(int i=front[x];i;i=nxt[i])
	{
		t=to[i];
		if(t==fa) continue;
		if(search(x,t)) continue;
		ch++;
	}
	if(ch<=1) return; 
	if(ch==2)
	{
		if(havefa) 
		{
			cutdown(fa,x);
		}
		return;
	}
	int now=0;
	if(havefa)
	{
		cutdown(fa,x);
	}
	for(int i=front[x];i;i=nxt[i]) 
	{	
		t=to[i];
		if(t==fa) continue;
		if(search(x,t)) continue;
		now++;
		if(now<=ch-2)
		{
			cutdown(x,t);
		}
	}
}

void findend(int x,int fa)
{
	en=x;
	vis[x]=true;
	int t;
	for(int i=front2[x];i;i=nxt2[i])
	{
		t=to2[i];
		if(t==fa) continue;
		findend(t,x);
	}
}

int findch(int x)
{
	return d[x];
}

void build()
{
	tot2=0;
	//memset(front2,0,sizeof(front2));
	for(int i=1;i<=n;++i) front2[i]=0;
	int u,v;
	for(int i=1;i<=n;++i)
		for(int j=front[i];j;j=nxt[j])
		{
			u=i;
			v=to[j];
			if(u>v) continue;
			if(edge.find(u*mod+v)==edge.end()) 
			{
				add2(u,v);
				add2(v,u);
			}
		}
	//memset(vis,false,sizeof(vis));
	for(int i=1;i<=n;++i) vis[i]=false;
	int tmp=0,ss;
	for(int i=1;i<=n;++i)
	{
		ss=findch(i);
		if(ss!=1 && ss!=0) continue;
		u=i;
		if(!vis[u])
		{
			findend(u,0);
			con[++tmp][0]=u;
			con[tmp][1]=en;
		}
	}
	printf("%d\n",ans);
	for(int i=1;i<ans;++i) printf("%d %d %d %d\n",cut[i][0],cut[i][1],con[i][0],con[i+1][1]);
	if(ans) printf("%d %d %d %d\n",cut[ans][0],cut[ans][1],con[ans][0],con[ans+1][0]);
}

int main()
{
	int T,u,v;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		tot=0;
		//memset(front,0,sizeof(front));
		for(int i=1;i<=n;++i) front[i]=0;
		for(int i=1;i<=n;++i) d[i]=0;
		for(int i=1;i<n;++i)
		{
			scanf("%d%d",&u,&v);
			add(u,v);
			add(v,u);
		}
		ans=0;
		edge.clear();
		dfs(1,0,0);
		build();
	}
}
