Problem: Snuke has found two trees A and B, each with N vertices numbered 1 to N. The i-th edge of A connects Vertex a_i and b_i, and the j-th edge of B connects Vertex c_j and d_j. Also, all vertices of A are initially painted white.

Snuke would like to perform the following operation on A zero or more times so that A coincides with B:

* Choose a leaf vertex that is painted white. (Let this vertex be v.)
* Remove the edge incident to v, and add a new edge that connects v to another vertex.
* Paint v black.



Determine if A can be made to coincide with B, ignoring color. If the answer is yes, find the minimum number of operations required.

You are given T cases of this kind. Find the answer for each of them.

Constraints

* 1 \leq T \leq 20
* 3 \leq N \leq 50
* 1 \leq a_i,b_i,c_i,d_i \leq N
* All given graphs are trees.

Input

Input is given from Standard Input in the following format:


T
case_1
:
case_{T}


Each case is given in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}
c_1 d_1
:
c_{N-1} d_{N-1}


Output

For each case, if A can be made to coincide with B ignoring color, print the minimum number of operations required, and print `-1` if it cannot.

Examples

Input

2
3
1 2
2 3
1 3
3 2
6
1 2
2 3
3 4
4 5
5 6
1 2
2 4
4 3
3 5
5 6


Output

1
-1


Input

3
8
2 7
4 8
8 6
7 1
7 3
5 7
7 8
4 2
5 2
1 2
8 1
3 2
2 6
2 7
4
1 2
2 3
3 4
3 4
2 1
3 2
9
5 3
4 3
9 3
6 8
2 3
1 3
3 8
1 7
4 1
2 8
9 6
3 6
3 5
1 8
9 7
1 6


Output

6
0
7

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
 
const int N = 55;
 
int n, par[N], e[N][N], te[N][N], deg[N], tdeg[N];
vector <int> gb[N];
 
void dfs(int u) {
	for (int v : gb[u]) if (v != par[u]) {
		par[v] = u;
		dfs(v);
	}
}
 
void change(vector <int> &available, int u) {
	available.erase(find(available.begin(), available.end(), u));
	for (auto v : gb[u]) if (v != par[u]) {
		available.push_back(v);
	}
}
 
int calc(int root) {
	par[root] = 0;
	dfs(root);
 
	for (int i = 1; i <= n; ++i) {
		tdeg[i] = deg[i];
		for (int j = 1; j <= n; ++j) {
			te[i][j] = e[i][j];
		}
	}
 
	vector <int> available(1, root);
	change(available, root);
 
	int ret = 0;
	while(!available.empty()) {
		bool found = false;
		for (auto u : available) {
			if (te[u][par[u]]) {
				found = true;
				change(available, u);
				break;
			}
			if (tdeg[u] == 1) { // leaf
				found = true;
				// cut
				for (int i = 1; i <= n; ++i) if (te[i][u]) {
					te[i][u] = te[u][i] = false;
					--tdeg[i]; --tdeg[u];
				}
				// concatenate
				te[u][par[u]] = te[par[u]][u] = true;
				++tdeg[u]; ++tdeg[par[u]];
				++ret;
 
				change(available, u);
				break;
			}
		}
		if (!found) return 1e9;
	}
 
	return ret;
}
 
void solve() {
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		deg[i] = 0;
		gb[i].clear();
		for (int j = 1; j <= n; ++j) {
			e[i][j] = 0;
		}
	}
 
	for (int i = 1; i < n; ++i) {
		int u, v; cin >> u >> v;
		e[u][v] = e[v][u] = true;
		++deg[u];
		++deg[v];
	}
 
	for (int i = 1; i < n; ++i) {
		int u, v; cin >> u >> v;
		gb[u].push_back(v);
		gb[v].push_back(u);
	}
 
	int res = 1e9;
	for (int i = 1; i <= n; ++i) {
		int cur = calc(i);
		res = min(res, cur);
	}
 
	for (int i = 1; i <= n; ++i) if (deg[i] == 1) {
		int adj_i = 0;
		for (int j = 1; j <= n; ++j) if (e[i][j]) {
			adj_i = j;
			e[i][j] = e[j][i] = false;
			--deg[i]; --deg[j];
		}
		for (int j = 1; j <= n; ++j) if (i != j) {
			e[i][j] = e[j][i] = true;
			++deg[i]; ++deg[j];
			res = min(res, calc(i) + 1);
			e[i][j] = e[j][i] = false;
			--deg[i]; --deg[j];
		}
		e[i][adj_i] = e[adj_i][i] = true;
		++deg[i]; ++deg[adj_i];
	}
 
	cout << (res == 1e9 ? -1 : res) << endl;
}
 
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
 
	int tt; cin >> tt;
	while(tt--) {
		solve();
	}
}