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; 
int gi() { 
	int res = 0, w = 1; 
	char ch = getchar(); 
	while (ch != '-' && !isdigit(ch)) ch = getchar(); 
	if (ch == '-') w = -1, ch = getchar(); 
	while (isdigit(ch)) res = res * 10 + ch - '0', ch = getchar(); 
	return res * w; 
}
const int INF = 1e9; 
const int MAX_N = 105;
int N; 
struct Graph { 
	vector<int> E[MAX_N]; 
	int fa[MAX_N], deg[MAX_N]; 
	void init() { for (int i = 1; i <= N; i++) fa[i] = deg[i] = 0, E[i].clear(); } 
	void Add_Edge(int u, int v) { E[u].push_back(v), ++deg[v]; } 
	void dfs(int x) { 
		for (int v : E[x]) 
			if (v != fa[x]) fa[v] = x, dfs(v); 
	} 
} A, B, C; 
bool check(int x) { return A.fa[x] == B.fa[x]; } 
int solve() { 
	int res = 0; 
	for (int i = 1; i <= N; i++) res += !check(i); 
	C.init(); 
	for (int i = 1; i <= N; i++) { 
		if (check(i)) {
			if (!check(A.fa[i])) return INF; 
			else continue; 
		} 
		if (!check(A.fa[i])) C.Add_Edge(i, A.fa[i]); 
		if (!check(B.fa[i])) C.Add_Edge(B.fa[i], i); 
	} 
	queue<int> que; int siz = 0;
	for (int i = 1; i <= N; i++) if (!C.deg[i]) que.push(i); 
	while (!que.empty()) {
		int x = que.front(); que.pop(), ++siz; 
		for (int v : C.E[x]) 
			if (!--C.deg[v]) que.push(v); 
	} 
	return siz == N ? res : INF; 
} 
int main () { 
	int T = gi(); 
	while (T--) { 
		N = gi(); 
		for (int i = 1; i < N; i++) { 
			int u = gi(), v = gi(); 
			A.Add_Edge(u, v), A.Add_Edge(v, u); 
		} 
		for (int i = 1; i < N; i++) { 
			int u = gi(), v = gi(); 
			B.Add_Edge(u, v), B.Add_Edge(v, u); 
		} 
		int ans = INF; 
		for (int i = 1; i <= N; i++) { 
			A.fa[i] = B.fa[i] = 0; 
			A.dfs(i), B.dfs(i), ans = min(ans, solve()); 
			if (A.deg[i] != 1) continue; 
			for (int j = 1; j <= N; j++) {
				if (i == j) continue; 
				A.fa[j] = i, A.dfs(j), A.fa[i] = 0, ans = min(ans, solve() + 1); 
			} 
		} 
		printf("%d\n", ans == INF ? -1 : ans); 
		A.init(), B.init(), C.init(); 
	} 
    return 0; 
}