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>
#define clear(x) memset(x, 0, sizeof(x))
using namespace std;

const int maxn = 50;
int T, n, a[maxn + 3][maxn + 3], b[maxn + 3][maxn + 3], c[maxn + 3], d[maxn + 3], deg[maxn + 3];
bool vis[maxn + 3];
vector<int> G[maxn + 3];

void add(int u, int v) {
	G[u].push_back(v), deg[v]++;
}

void dfs(int u) {
	vis[u] = true;
	for (int v = 1; v <= n; v++) {
		if (a[u][v] && b[u][v] && !vis[v]) {
			dfs(v);
		}
	}
}

void dfs_1(int u, int pa = 0) {
	for (int v = 1; v <= n; v++) {
		if (a[u][v] && v != pa) {
			if (!vis[u]) {
				add(v, u);
			}
			dfs_1(v, u);
		}
	}
}

void dfs_2(int u, int pa = 0) {
	for (int v = 1; v <= n; v++) {
		if (b[u][v] && v != pa) {
			if (!vis[u]) {
				add(u, v);
			}
			dfs_2(v, u);
		}
	}
}

bool check() {
	queue<int> Q;
	int cnt = 0;
	for (int i = 1; i <= n; i++) {
		if (!deg[i]) {
			Q.push(i);
		}
	}
	while (!Q.empty()) {
		int u = Q.front();
		Q.pop(), cnt++;
		for (int i = 0, v; i < G[u].size(); i++) {
			v = G[u][i];
			if (!--deg[v]) {
				Q.push(v);
			}
		}
	}
	return cnt == n;
}

int solve(int u) {
	clear(vis);
	dfs(u);
	for (int i = 1; i <= n; i++) {
		vector<int>().swap(G[i]);
	}
	clear(deg);
	dfs_1(u);
	dfs_2(u);
	if (!check()) {
		return -1;
	}
	int cnt = 0;
	for (int i = 1; i <= n; i++) {
		cnt += vis[i];
	}
	return cnt;
}

int main() {
	scanf("%d", &T);
	while (T --> 0) {
		scanf("%d", &n);
		clear(a), clear(b), clear(c), clear(d);
		for (int i = 1, u, v; i < n; i++) {
			scanf("%d %d", &u, &v);
			a[u][v] = a[v][u] = 1;
			c[u]++, c[v]++;
		}
		for (int i = 1, u, v; i < n; i++) {
			scanf("%d %d", &u, &v);
			b[u][v] = b[v][u] = 1;
			d[u]++, d[v]++;
		}
		int ans = -1;
		for (int i = 1; i <= n; i++) {
			ans = max(ans, solve(i));
		}
		for (int i = 1; i <= n; i++) if (c[i] == 1) {
			for (int j = 1; j <= n; j++) if (a[i][j]) {
				a[i][j] = a[j][i] = 0;
				for (int k = 1; k <= n; k++) if (k != i && k != j) {
					a[i][k] = a[k][i] = 1;
					ans = max(ans, solve(i) - 1);
					a[i][k] = a[k][i] = 0;
				}
				a[i][j] = a[j][i] = 1;
			}
		}
		printf("%d\n", ans == -1 ? -1 : n - ans);
	}
	return 0;
}