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 ll int
#define f first
#define sc second
#define pb push_back
using namespace std;
ll a,b,c,d,i,e,f,g,n,m,k,l,t;
ll del[200005],fix[200005],fix1[200005];
pair <ll,ll> B[200005];
vector < pair <ll,ll> > v[200005],vans1,vans2;
vector <ll> v1;
void dfs(ll x,ll y) {
    int k=v[x].size();
    if(x!=1) k--;
    for(ll i=0;i<v[x].size();i++) {
        if(v[x][i].f==y) continue;
        dfs(v[x][i].f,x);
        if(fix[v[x][i].f]) { k--; del[v[x][i].sc]=1; }
    }
    if(k>=2) {
        fix[x]=1;
        for(ll i=0;i<v[x].size();i++) {
            if(v[x][i].f==y) continue;
            if(k==2) break;
            if(fix[v[x][i].f]==0) { k--; del[v[x][i].sc]=1; }
        }
    }
}
void dfs1(ll x,ll y) {
    fix1[x]=1;
    int k=0;
    for(ll i=0;i<v[x].size();i++) {
        if(fix1[v[x][i].f]==1 || del[v[x][i].sc]==1) continue;
        k++;
        dfs1(v[x][i].f,y);
    }
    if(k==0) v1.pb(x);
    else if(x==y && k==1) v1.pb(x);
}
int main() {
    cin>>t;
    while(t--) {
        cin>>n;
        for(ll i=1;i<n;i++) {
            cin>>a>>b;
            v[a].pb({b,i});
            v[b].pb({a,i});
        }
        dfs(1,0);
        l=0;
        for(ll i=1;i<=n;i++) {
            if(fix1[i]==0) {
                dfs1(i,i); 
                l++;
                //if(i==2)cout<<v1.size()<<endl;
                if(v1.size()==1) { B[l].f=v1[0]; B[l].sc=v1[0]; }
                else { B[l].f=v1[0]; B[l].sc=v1[1]; }
               // if(i==2) cout<<B[l].f<<" "<<B[l].sc<<endl;
                v1.clear();
            }
        }
        vans1.clear(); vans2.clear();
        for(ll i=1;i<=n;i++) {
            for(ll j=0;j<v[i].size();j++) {
                if(del[v[i][j].sc]==1) {
                    if(i<v[i][j].f) { vans1.pb( { i, v[i][j].f } ); }
                }
            }
        }
        for(ll i=2;i<=l;i++) {
            vans2.pb( { B[i].f, B[i-1].sc } );
        }
        cout<<vans1.size()<<endl;
        for(ll i=0;i<vans1.size();i++) {
            cout<<vans1[i].f<<" "<<vans1[i].sc<<" "<<vans2[i].f<<" "<<vans2[i].sc<<endl;
        }
        for(ll i=1;i<=n;i++) {
            del[i]=0; fix[i]=0; fix1[i]=0; v[i].clear();
        }
    }
}