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;
typedef long long ll;
ll t,n,m,x,y,u,v,cnt,timer,pos,vis[100009],pa[100009],id[100009],lft[100009],rgt[100009],wtf[100009];
vector <ll> to[100009];
struct st{ll a,b,c,d;}p[100009];
ll find(ll x){return x == id[x] ? x : id[x] = find(id[x]);}
inline ll read(){
	ll s = 0,w = 1;
	char ch = getchar();
	while (ch > '9' || ch < '0'){ if (ch == '-') w = -1; ch = getchar();}
	while (ch <= '9' && ch >= '0') s = (s << 1) + (s << 3) + (ch ^ 48),ch = getchar();
	return s * w;
}
void dfs1(ll u,ll fa){
	pa[u] = fa,lft[u] = rgt[u] = u;
	for (ll i = 0;i < to[u].size();i += 1){
		ll v = to[u][i];
		if (v == fa) continue;
		dfs1(v,u);
	}
}
void dfs2(ll u,ll fa){
	for (ll i = 0;i < to[u].size();i += 1){
		ll v = to[u][i];
		if (v == fa) continue;
		dfs2(v,u);
		if (!vis[v]){
			if (lft[u] == u) id[find(v)] = find(u),lft[u] = lft[v],wtf[v] = 1;
			else if (rgt[u] == u) id[find(v)] = find(u),rgt[u] = lft[v],vis[u] = 1,wtf[v] = 2;
		}
	}
}
void dfs3(ll u,ll fa){
	for (ll i = 0;i < to[u].size();i += 1){
		ll v = to[u][i];
		if (v == fa) continue;
		dfs3(v,u);
	}
	if (id[u] == u && u > 1){
		if (wtf[pa[u]] == 1) p[++ cnt] = {u,pa[u],lft[u],lft[id[pa[u]]]},lft[id[pa[u]]] = rgt[u];
		else p[++ cnt] = {u,pa[u],lft[u],rgt[id[pa[u]]]},rgt[id[pa[u]]] = rgt[u];
	}
}
int main(){
	t = read();
	while (t --){
		n = read(),cnt = 0;
		for (ll i = 1;i <= n;i += 1) to[i].clear(),vis[i] = 0,id[i] = i;
		for (ll i = 1;i < n;i += 1) u = read(),v = read(),to[u].push_back(v),to[v].push_back(u);
		dfs1(1,0),dfs2(1,0);
		for (ll i = 1;i <= n;i += 1) id[i] = find(i);
		dfs3(1,0);
		printf("%lld\n",cnt);
		for (ll i = 1;i <= cnt;i += 1) printf("%lld %lld %lld %lld\n",p[i].a,p[i].b,p[i].c,p[i].d);
	}
	return 0;
}