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;

namespace TYC
{
	const int N = 55, INF = 0x3f3f3f3f;

	int n, ans, FA[N], FB[N], Deg[N];
	vector<int> A[N], B[N], E[N];

	inline int read()
	{
		int x = 0, f = 0, ch = getchar();
		while (!isdigit(ch))
			f |= (ch == '-'), ch = getchar();
		while (isdigit(ch))
			x = x * 10 + ch - '0', ch = getchar();
		return f ? -x : x;
	}

	void dfs(const int u, const int f, vector<int> *T, int *fa)
	{
		fa[u] = f;
		for (int v : T[u])
			if (v != f)
				dfs(v, u, T, fa);
	}

	bool toposort(int tot)
	{
		static queue<int> q;
		for (int i = 1; i <= n; i++)
			if (FA[i] != FB[i] && !Deg[i])
				q.push(i);
		while (!q.empty())
		{
			int u = q.front();
			q.pop();
			tot--;
			for (int v : E[u])
				if (!--Deg[v])
					q.push(v);
		}
		return !tot;
	}

	inline int count(const int u, const int f)
	{
		int c = 1;
		for (int v : A[u])
			if (v != f && FB[v] == u)
				c += count(v, u);
		return c;
	}

	int solve(const int o)
	{
		dfs(o, 0, A, FA);
		dfs(o, 0, B, FB);
		for (int i = 1; i <= n; i++)
			E[i].clear(), Deg[i] = 0;
		int cnt = 0;
		for (int i = 1, a, b; i <= n; i++)
			if ((a = FA[i]) != (b = FB[i]))
			{
				cnt++;
				if (FA[a] != FB[a])
					E[i].push_back(a), Deg[a]++;
				if (FA[b] != FB[b])
					E[b].push_back(i), Deg[i]++;
			}
		return count(o, 0) == n - cnt && toposort(cnt) ? cnt : INF;  
	}

	void work()
	{
		int T = read();
		while (T--)
		{
			n = read();
			for (int i = 1; i < n; i++)
			{
				int u = read(), v = read();
				A[u].push_back(v);
				A[v].push_back(u);
			}
			for (int i = 1; i < n; i++)
			{
				int u = read(), v = read();
				B[u].push_back(v);
				B[v].push_back(u);
			}
			ans = INF;
			for (int i = 1; i <= n; i++)
				ans = min(ans, solve(i));
			for (int u = 1; u <= n; u++)
				if (int(A[u].size()) == 1)
				{
					int v = A[u][0];
					A[u].pop_back();
					A[v].erase(find(A[v].begin(), A[v].end(), u));
					for (int p = 1; p <= n; p++)
						if (p != u && p != v)
						{
							A[u].push_back(p);
							A[p].push_back(u);
							ans = min(ans, solve(u) + 1);
							A[u].pop_back();
							A[p].pop_back();
						}
					A[u].push_back(v);
					A[v].push_back(u);
				}
			printf("%d\n", ans == INF ? -1 : ans);
			for (int i = 1; i <= n; i++)
				A[i].clear(), B[i].clear();
		}
	}
}

int main()
{
//	freopen("4380.in", "r", stdin);
	TYC::work();
	return 0;
}