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 MX = 50;

namespace dsu {
	int p[MX];
	
	void init(int n) {
		iota(p, p + n, 0);
	}
	
	int get(int v) {
		if (p[v] != v) p[v] = get(p[v]);
		return p[v];
	}
	
	bool merge(int u, int v) {
		u = get(u);
		v = get(v);
		p[u] = v;
		return u != v;
	}
}

int n, par[MX];
vector<int> G[MX];
vector<pair<int, int>> edges;
bool enabled[MX];

void dfs(int v, int p = -1, bool enable = true) {
	par[v] = p;
	enabled[v] = enable;
	for (int u : G[v]) {
		if (u == p) continue;
		if (enable && dsu::get(u) == dsu::get(v))
			dfs(u, v, true);
		else
			dfs(u, v, false);
	}
}

int solve(int root) {
	dfs(root);
	vector<int> deg(n, 0);
	vector<int> GG[MX];
	for (auto& e : edges) {
		deg[e.first]++;
		deg[e.second]++;
		
		GG[e.first].push_back(e.second);
		GG[e.second].push_back(e.first);
	}
	
	int res = 0;
	bool ok = true;
	while (ok) {
		ok = false;
		for (int i = 0; i < n; i++)
			if (deg[i] == 1 && enabled[i] == false && enabled[par[i]]) {
				enabled[i] = true;
				res++;
				for (int u : GG[i]) {
					deg[u]--;
				}
				ok = true;
			}
	}
	
	for (int i = 0; i < n; i++) if (enabled[i] == false) return n + 1;
	return res;
}

int main() {
	int T;
	ignore = scanf("%d", &T);
	while (T--) {
		ignore = scanf("%d", &n);
		
		edges.clear();
		dsu::init(n);
		for (int i = 0; i < n; i++) {
			G[i].clear();
		}
		
		for (int i = 1, u, v; i < n; i++) {
			ignore = scanf("%d %d", &u, &v);
			u--; v--;
			edges.emplace_back(u, v);
		}
		
		for (int i = 1, u, v; i < n; i++) {
			ignore = scanf("%d %d", &u, &v);
			u--; v--;
			G[u].emplace_back(v);
			G[v].emplace_back(u);
			
			for (auto& e : edges)
				if ((e.first == u && e.second == v) || (e.first == v && e.second == u)) {
					dsu::merge(u, v);
				}
		}
		
		int ans = n + 1;
		for (int i = 0; i < n; i++) ans = min(solve(i), ans);
		
		if (ans > n) {
			dsu::init(n);
			for (int v = 0; v < n; v++) {
				int id = -1;
				for (int i = 0; i < n - 1; i++)
					if (edges[i].first == v || edges[i].second == v) {
						if (id != -1) id = -2;
						else id = i;
					}
				
				if (id >= 0) {
					auto e = edges[id];
					for (int u = 0; u < n; u++) {
						if (u == v) continue;
						edges[id] = make_pair(v, u);
						if (solve(v) != n + 1) ans = n;
					}
					edges[id] = e;
				}
			}
		}
		
		if (ans > n) ans = -1;
		printf("%d\n", ans);
	}
	
	return 0;
}