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 <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cassert>
#include <ctime>
#include <map>
#include <queue>
#include <set>
#include <cmath>
#include <list>
using namespace std;


#define N 100

int n;
int fa[N], fa2[N], deg[N];
vector <int> ve[N], ve2[N];
bool ok[N];

void dfs2(int k, int f) {
	fa2[k] = f;
	for (int i = 0; i < (int) ve2[k].size(); i++)
		if (ve2[k][i] != f) {
			dfs2(ve2[k][i], k);
		}
}

void dfs(int k, int f) {
	fa[k] = f;
	for (int i = 0; i < (int) ve[k].size(); i++)
		if (ve[k][i] != f) {
			dfs(ve[k][i], k);
		}
}

int calc(int r) {
	for (int i = 1; i <= n; i++)
		ok[i] = false;
	dfs2(r, 0);
	dfs(r, 0);

	ok[r] = true;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++)
			if (ok[fa[j]] && fa[j] == fa2[j])
				ok[j] = true;
	}	
	for (int i = 1; i <= n; i++)
		deg[i] = 0;
	for (int i = 1; i <= n; i++)
		deg[fa[i]]++;
	int op = 0;
	for (int i = 1; i <= n; i++) {
		int k = -1;
		for (int j = 1; j <= n; j++)
			if (!ok[j] && !deg[j] && ok[fa2[j]]) {
				k = j;
				break;
			}
		if (k == -1) {
			break;
		}
		op++;
		ok[k] = true;
		deg[fa[k]]--;
	}
	for (int i = 1; i <= n; i++)
	if (!ok[i])
		return 1e9;
	return op;
}


int calc2(int r1, int r2) {
	for (int i = 1; i <= n; i++)
		ok[i] = false;
	dfs2(r2, 0);
	dfs(r1, 0);

	for (int i = 1; i <= n; i++)
		deg[i] = 0;
	for (int i = 1; i <= n; i++)
		deg[fa[i]]++;
	int op = 0;
	ok[0] = true;
	for (int i = 1; i <= n; i++) {
		int k = -1;
		for (int j = 1; j <= n; j++)
			if (!ok[j] && !deg[j] && ok[fa2[j]]) {
				k = j;
				break;
			}
		if (k == -1) {
			break;
		}
		op++;
		ok[k] = true;
		deg[fa[k]]--;
	}
	for (int i = 1; i <= n; i++)
	if (!ok[i])
		return 1e9;
	return op;
}

int main() {
	int T;
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
			ve[i].clear(), ve2[i].clear();
		for (int i = 1; i < n; i++) {
			int a, b;
			scanf("%d%d", &a, &b);
			ve[a].push_back(b);
			ve[b].push_back(a);
		}
		for (int i = 1; i < n; i++) {
			int a, b;
			scanf("%d%d", &a, &b);
			ve2[a].push_back(b);
			ve2[b].push_back(a);
		}
		int ans = 1e9;
		for (int root = 1; root <= n; root++) {
			ans = min(ans, calc(root));
		}
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				ans = min(ans, calc2(i, j));

		printf("%d\n", ans == 1000000000 ? -1 : ans);
	}
}