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;
int t,n,a[200007],b[200007],tim,tt;
int dp[200007];
bool fr[200007];
int used[200007],visited[200007];
vector< pair<int,int> > v[200007];
vector< pair<int,int> > ans,ans2;
pair<int,int> p[200007];
void dfs(int x,int p){
    fr[x]=true;
    dp[x]=0;
    int cnt=0;
    for(int i=0;i<v[x].size();i++){
        if(v[x][i].first!=p){
            dfs(v[x][i].first,x);
            if(fr[v[x][i].first])cnt++;
            dp[x]+=dp[v[x][i].first];
        }
    }
    if(cnt==0)dp[x]++;
    if(cnt>=2){dp[x]--;fr[x]=false;}
    cnt=min(cnt,2);
    for(int i=0;i<v[x].size();i++){
        if(v[x][i].first!=p){
            if(cnt==0 or !fr[v[x][i].first]){ans.push_back(make_pair(x,v[x][i].first));used[v[x][i].second]=tim;}
            else {cnt--;}
        }
    }
}
void dfs2(int x){
    visited[x]=tim;
    if(v[x].size()==1){
        if(p[tt].first==-1)p[tt].first=x;
        else p[tt].second=x;
    }else if(v[x].size()==0){
        p[tt]=make_pair(x,x);
    }
    for(int i=0;i<v[x].size();i++){
        if(visited[v[x][i].first]!=tim)dfs2(v[x][i].first);
    }
}
int main(){
    cin>>t;
    for(int i=0;i<t;i++){
        cin>>n;
        tim++;tt=0;
        for(int f=1;f<=n;f++)v[f].clear();
        ans.clear();ans2.clear();
        for(int f=0;f<n-1;f++){
            cin>>a[f]>>b[f];
            v[a[f]].push_back(make_pair(b[f],f));
            v[b[f]].push_back(make_pair(a[f],f));
        }
        dfs(1,0);
        cout<<dp[1]-1<<"\n";
        for(int f=1;f<=n;f++)v[f].clear();
        for(int f=0;f<n-1;f++){
            if(used[f]!=tim){
                v[a[f]].push_back(make_pair(b[f],f));
                v[b[f]].push_back(make_pair(a[f],f));
            }
        }
        for(int f=1;f<=n;f++){
            if(visited[f]!=tim){
                p[tt]=make_pair(-1,-1);
                dfs2(f);
                tt++;
            }
        }
        for(int f=0;f<tt-1;f++){
            ans2.push_back(make_pair(p[f].second,p[f+1].first));
        }
        for(int f=0;f<ans.size();f++){
            cout<<ans[f].first<<" "<<ans[f].second<<" "<<ans2[f].first<<" "<<ans2[f].second<<"\n";
        }
    }
    return 0;
}