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 <cstdio>
#include <vector>
using namespace std;
#define maxn 100005
struct node {
	int v1, v2, v3, v4;
	node(){}
	node( int V1, int V2, int V3, int V4 ) {
		v1 = V1, v2 = V2, v3 = V3, v4 = V4;
	}
};
vector < node > ans;
vector < int > G[maxn];
int T, n;

int dfs( int u, int fa ) {
	int pos = u, cnt = 0;
	for( int i = 0;i < G[u].size();i ++ ) {
		int v = G[u][i];
		if( v == fa ) continue;
		int now = dfs( v, u );
		if( now < 0 ) continue;
		cnt ++;
		if( cnt == 1 ) pos = now;
		else if( cnt == 2 ) ans.push_back( node( u, fa, now, pos ) ), pos = -1;
		else ans.push_back( node( u, v, v, now ) );
	}
	return pos;
}

int main() {
	scanf( "%d", &T );
	while( T -- ) {
		scanf( "%d", &n );
		for( int i = 1;i <= n;i ++ )
			G[i].clear();
		ans.clear();
		for( int i = 1, u, v;i < n;i ++ ) {
			scanf( "%d %d", &u, &v );
			G[u].push_back( v );
			G[v].push_back( u );
		}
		int rt = 1;
		while( rt <= n && G[rt].size() != 1 ) rt ++;
		int head = dfs( rt, -1 );
		printf( "%d\n", ans.size() );
		for( int i = 0;i < ans.size();i ++ )
			printf( "%d %d %d %d\n", ans[i].v1, ans[i].v2, ans[i].v3, head ), head = ans[i].v4;
	}
	return 0;
}