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 fi first
#define se second
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (int)(x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

using namespace std;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;

const int INF = 0x3f3f3f3f;
const double PI = acos(-1);
const int N = 1e6+10;

vector<pii>e[N];
vector<pii>ans[2];

int nofa[N],del[N],vis[N];

void dfs1(int u,int fa) {
	int son = 0;
	for(auto x:e[u]) {
		int v = x.fi;
		if(v == fa) continue;
		dfs1(v,u);
		if(!nofa[v]) son++;
		else del[x.se] = 1;
	}
	if(son>=2) {
		nofa[u] = 1;
		for(auto x:e[u]) {
			int v = x.fi;
			if(v == fa) continue;
			if(son<=2) break;
			if(!nofa[v]) {
				son--;
				del[x.se] = 1;
			}
		}
	}
}

vector<int>leaf;

void dfs2(int u,int fa) {
	int son = 0;
	vis[u] = 1;
	for(auto x:e[u]) {
		int v = x.fi;
		if(v == fa || del[x.se]) continue;
		dfs2(v,u);
		son++;
	}
	if(fa == -1 && son == 1) leaf.pb(u);
	else if(son == 0) leaf.pb(u);
}

signed main() {

#ifdef xiaofan
	freopen("1.in","r",stdin);
	freopen("1.out","w",stdout);
#endif

	int T;
	cin>>T;
	while(T--) {
		int n;
		cin>>n;
		ans[1].clear();
		ans[0].clear();
		for(int i=1; i<=n; i++) {
			vis[i] = 0;
			del[i] = 0;
			nofa[i] = 0;

			e[i].clear();
		}
		for(int i=1; i<n; i++) {
			int u,v;
			cin>>u>>v;
			e[u].pb(mp(v,i));
			e[v].pb(mp(u,i));
		}
		dfs1(1,-1);
		for(int i=1; i<=n; i++) {
			for(auto x:e[i]) {
				if(del[x.se] && i<x.fi) {
					ans[0].pb(mp(i,x.fi));
				}
			}
		}
		vector<pii>tmp;
		for(int i=1; i<=n; i++) {
			if(!vis[i]) {
				leaf.clear();
				dfs2(i,-1);
				if(sz(leaf) == 2) {
					tmp.pb(mp(leaf[0],leaf[1]));
				} else {
					tmp.pb(mp(leaf[0],leaf[0]));
				}
			}
		}
		for(int i=0; i<(int)sz(tmp)-1; i++) ans[1].pb(mp(tmp[i].se,tmp[i+1].fi));
		cout<<sz(ans[0])<<endl;
		for(int i=0; i<sz(ans[0]); i++) cout<<ans[0][i].fi<<" "<<ans[0][i].se<<" "<<ans[1][i].fi<<" "<<ans[1][i].se<<endl;




	}



	return 0;
}