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;

// #define int long long
#define pb push_back
#define all(v) begin(v), end(v)
#define sz(v) (int)v.size()
#define ff first
#define ss second
const int inf = 4e18;
const int N = 2e5 + 10;

vector<pair<int, int>> del;
vector<set<int>> adjList;
vector<vector<int>> to_rem;

void dfs(int nd, int par) {
	for (int i : adjList[nd]) {
		if (i != par) dfs(i, nd);
	}

	for (int i : to_rem[nd])
		adjList[nd].erase(i);

	int ch = adjList[nd].size() - 1;
	if (ch >= 2 and nd != par) {
		adjList[nd].erase(par);
		to_rem[par].pb(nd);
		del.pb({par, nd});
	}
	if (ch > 2) {
		vector<int> rem;
		int k = ch - 2;
		for (int i : adjList[nd]) {
			if (i == par) continue;
			if (!k) break;
			rem.pb(i);
			k--;
		}

		for (int &i : rem) {
			adjList[nd].erase(i);
			adjList[i].erase(nd);
			del.pb({nd, i});
		}
	}
}

void dfs2(int nd, int par, vector<int> &lf, vector<int> &vis, int cur) {
	vis[nd] = cur;
	if (adjList[nd].size() <= 1) lf.pb(nd);

	for (int i : adjList[nd]) {
		if (i != par) dfs2(i, nd, lf, vis, cur);
	}
}

void testCase() {
	int n; cin >> n;
	adjList = vector<set<int>>(n + 1);
	to_rem = vector<vector<int>>(n + 1);
	del.clear();

	for (int i = 1; i < n; i++) {
		int u, v; cin >> u >> v;
		adjList[u].insert(v);
		adjList[v].insert(u);
	}

	adjList[1].insert(1);
	dfs(1, 1);
	vector<int> vis(n + 1, -1);
	vector<vector<int>> leafs;
	int cur = 0;
	for (int i = 1; i <= n; i++) {
		if (vis[i] == -1) {
			vector<int> lf;
			dfs2(i, i, lf, vis, cur++);
			if (lf.size() == 1) lf.pb(lf.back());
			leafs.pb(lf);
		}
	}

	assert(leafs.size() == del.size() + 1);

	cout << del.size() << '\n';
	for (int i = 0; i < (int)del.size(); i++) {
		int u = del[i].ff, v = del[i].ss;
		int u1 = leafs[i][1], v1 = leafs[i + 1][0];
		cout << u << ' ' << v << ' ' << u1 << ' ' << v1 << '\n';
	}
	cout << endl;
}

int32_t main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int t_c = 1;
	cin >> t_c;
	while (t_c--) testCase();
}