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;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    vector<vector<int>> g(n);
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      g[a - 1].push_back(b - 1);
      g[b - 1].push_back(a - 1);
    }
    vector<tuple<int, int, int, int>> ans;
    vector<int> tail(n), head(n);
    function<void(int, int)> dfs = [&](int a, int p) {
        tail[a] = head[a] = a;
        for (int b : g[a]) {
          if (b != p) {
            dfs(b, a);
            if (head[b] == b) {
              if (head[a] == a) {
                if (tail[a] == a) tail[a] = tail[b]; else head[a] = tail[b];
              } else {
                ans.emplace_back(a, b, tail[a], b);
                tail[a] = tail[b];
              }
            }
          }
        }
        for (int b : g[a]) {
          if (b != p && head[b] != b) {
            ans.emplace_back(a, b, tail[a], head[b]);
            tail[a] = tail[b];
          }
        }
    };
    dfs(0, -1);
    cout << ans.size() << '\n';
    for (auto [x1, y1, x2, y2] : ans)
      cout << x1 + 1 << ' ' << y1 + 1 << ' ' << x2 + 1 << ' ' << y2 + 1 << '\n';
  }
  return 0;
}

// Why is a raven like a writing desk?