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;
void read (int &x) {
    char ch = getchar(); int f = 0; x = 0;
    while (!isdigit(ch)) { if (ch == '-') f = 1; ch = getchar(); }
    while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
    if (f) x = -x;
} const int N = 2e5 + 5;
int n, d[N], fa[N], c[N], f[N], vis[N]; vector<int> g[N], t[N];
#define pb push_back
void dfs (int u, int la) {
    d[u] = d[la] + 1, fa[u] = la;
    for (int v : g[u]) if (v != la) dfs (v, u);
}
pair<int, int> p[N]; int co[N];
#define fi first
#define se second
int get (int x) { return f[x] == x ? x : f[x] = get (f[x]); }
struct pp {
    int x, y, lca, d;
    bool operator < (const pp &x) const {
        return d < x.d;
    }
} o[N]; int num;
struct res { int xa, ya, xb, yb;  } q[N]; int ans;
signed main() {
    int T; read (T);
    while (T--) {
        read (n); num = ans = 0;
        // g vis f c  t co
        for (int i = 1; i <= n; ++i) f[i] = i, vis[i] = co[i] = 0;
        for (int i = 1; i <= n; ++i) g[i].clear (), c[i] = 0, t[i].clear();
        for (int i = 1, u, v; i < n; ++i)
            read (u), read (v), g[u].pb (v), g[v].pb (u);
        d[0] = 0; dfs (1, 0);
        for (int i = 1; i <= n; ++i) p[i] = {-d[i], i};
        sort (p + 1, p + n + 1);
        for (int j = 1; j <= n; ++j) {
            int u = p[j].se;
            for (int v : g[u]) {
                if (c[u] == 2) break;
                if (vis[v] || v == fa[u]) continue;
                int fx = get (u), fy = get (v);
                if (fx != fy) f[fx] = fy, ++c[u];
            }
            if (c[u] == 2) vis[u] = 1;
        }
        for (int i = 1; i <= n; ++i) t[get (i)].pb (i), co[i] = get (i);

        // for (int i = 1; i <= n; ++i) {
        //     printf ("666\n");
        //     for (int j : t[i]) printf ("%d ", j); puts ("");
        // }
        d[0] = 1e9;
        for (int i = 1; i <= n; ++i) {
            // if (!(int)t[i].size()) continue;
            int a = 0, b = 0, lca = 0;
            for (int j : t[i]) {
                int ss = 0;
                // printf ("123 %d %d\n", d[j], d[lca]);
                if (d[j] < d[lca]) lca = j;
                for (int k : g[j]) ss += (co[k] == i);
                if (ss < 2) {
                    if (!a) a = j; else b = j;
                }
            }
            // cout << lca << endl;
            if (!b) b = a;
            if (!lca) continue;
            o[++num] = {a, b, lca, d[lca]};
        }
        // cout << num << endl;
        sort (o + 1, o + num + 1);
        int now = o[1].x;
        for (int i = 2; i <= num; ++i) {
            q[++ans] = {o[i].lca, fa[o[i].lca], now, o[i].y};
            now = o[i].x;
        }
        printf ("%d\n", ans);
        for (int i = 1; i <= ans; ++i) printf ("%d %d %d %d\n", q[i].xa, q[i].ya, q[i].xb, q[i].yb);
    }
}