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 <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
vector < int > edge[N];
vector < tuple <int, int, int, int> > ans;
int md;
int dfs(int x, int f) {
	int ch = 0, t = x;
	for (int u:edge[x]) {
		if (u == f)	continue;
		int v = dfs(u, x);
		if (!v)	continue;
		ch++;
		if (ch == 1)	t = v;
		else if (ch == 2)	ans.emplace_back(x, f, v, md), md = t;
		else ans.emplace_back(u, x, u, md), md = v;	
	}	
	return (ch <= 1) ? t : 0; 
}
int main() {
	int T;
	scanf("%d", &T);
	while (T--) {
		int n;
		scanf("%d", &n);
		for (int i = 1; i < n; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			edge[x].push_back(y);
			edge[y].push_back(x);
		}
		md = 1;
		while (edge[md].size() != 1)	md++;
		dfs(md, 0);
		printf("%d\n", ans.size());
		for (auto [a, b, c, d]:ans)
			printf("%d %d %d %d\n", a, b, c, d);
		ans.clear();
		for (int i = 1; i <= n; i++)	edge[i].clear();
	}
	return 0;
}