#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <vector>
using namespace std;

namespace zyt
{
	template<typename T>
	inline bool read(T &x)
	{
		char c;
		bool f = false;
		x = 0;
		do
			c = getchar();
		while (c != EOF && c != '-' && !isdigit(c));
		if (c == EOF)
			return false;
		if (c == '-')
			f = true, c = getchar();
		do
			x = x * 10 + c - '0', c = getchar();
		while (isdigit(c));
		if (f)
			x = -x;
		return true;
	}
	template<typename T>
	inline void write(T x)
	{
		static char buf[20];
		char *pos = buf;
		if (x < 0)
			putchar('-'), x = -x;
		do
			*pos++ = x % 10 + '0';
		while (x /= 10);
		while (pos > buf)
			putchar(*--pos);
	}
	typedef long long ll;
	const int N = 60, P = 1e9 + 7;
	int n, deg[N], fa[2][N];
	bool mp[2][N][N];
	vector<int> g[N];
	void dfs(const int id, const int u, const int f)
	{
		fa[id][u] = f;
		for (int i = 1; i <= n; i++)
			if (i != f && mp[id][u][i])
				dfs(id, i, u);
	}
	bool vis[N], insta[N];
	bool dfs2(const int u)
	{
		insta[u] = true;
		for (auto v : g[u])
			if (!vis[v] && (insta[v] || !dfs2(v)))
				return false;
		vis[u] = true, insta[u] = false;
		return true;
	}
	int solve(const int r)
	{
		int ans = 0;
		dfs(0, r, 0), dfs(1, r, 0);
		for (int i = 1; i <= n; i++)
			g[i].clear();
		for (int i = 1; i <= n; i++)
		{
			if (fa[0][i] != fa[1][i])
			{
				++ans;
				if (fa[0][fa[0][i]] != fa[1][fa[0][i]])
					g[i].push_back(fa[0][i]);
				if (fa[0][fa[1][i]] != fa[1][fa[1][i]])
					g[fa[1][i]].push_back(i);
			}
			else
			{
				int tmp = i;
				while (tmp)
				{
					if (fa[0][tmp] != fa[1][tmp])
						return n + 1;
					tmp = fa[0][tmp];
				}
				tmp = i;
				while (tmp)
				{
					if (fa[0][tmp] != fa[1][tmp])
						return n + 1;
					tmp = fa[1][tmp];
				}
			}
		}
		memset(vis, 0, sizeof(bool[n + 1]));
		memset(insta, 0, sizeof(bool[n + 1]));
		for (int i = 1; i <= n; i++)
			if (fa[0][i] != fa[1][i] && !vis[i] && !dfs2(i))
				return n + 1;
		return ans;
	}
	int work()
	{
		int T;
		read(T);
		while (T--)
		{
			read(n);
			int ans = n + 1;
			memset(mp, 0, sizeof(mp));
			memset(deg, 0, sizeof(int[n + 1]));
			for (int i = 0; i < 2; i++)
				for (int j = 1; j < n; j++)
				{
					int a, b;
					read(a), read(b);
					mp[i][a][b] = mp[i][b][a] = true;
				}
			for (int i = 1; i <= n; i++)
			{
				int d = 0, t = 0;
				for (int j = 1; j <= n; j++)
					if (mp[0][i][j])
						++d, t = j;
				if (d > 1)
					continue;
				for (int j = 1; j <= n; j++)
					if (i != j)
					{
						mp[0][i][t] = mp[0][t][i] = 0;
						mp[0][i][j] = mp[0][j][i] = 1;
						ans = min(ans, solve(i) + (t != j));
						mp[0][i][j] = mp[0][j][i] = 0;
						mp[0][i][t] = mp[0][t][i] = 1;
					}
			}
			write(ans > n ? -1 : ans), putchar('\n');
		}
		return 0;
	}
}
int main()
{
	return zyt::work();
}
### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <vector>
using namespace std;

namespace zyt
{
	template<typename T>
	inline bool read(T &x)
	{
		char c;
		bool f = false;
		x = 0;
		do
			c = getchar();
		while (c != EOF && c != '-' && !isdigit(c));
		if (c == EOF)
			return false;
		if (c == '-')
			f = true, c = getchar();
		do
			x = x * 10 + c - '0', c = getchar();
		while (isdigit(c));
		if (f)
			x = -x;
		return true;
	}
	template<typename T>
	inline void write(T x)
	{
		static char buf[20];
		char *pos = buf;
		if (x < 0)
			putchar('-'), x = -x;
		do
			*pos++ = x % 10 + '0';
		while (x /= 10);
		while (pos > buf)
			putchar(*--pos);
	}
	typedef long long ll;
	const int N = 60, P = 1e9 + 7;
	int n, deg[N], fa[2][N];
	bool mp[2][N][N];
	vector<int> g[N];
	void dfs(const int id, const int u, const int f)
	{
		fa[id][u] = f;
		for (int i = 1; i <= n; i++)
			if (i != f && mp[id][u][i])
				dfs(id, i, u);
	}
	bool vis[N], insta[N];
	bool dfs2(const int u)
	{
		insta[u] = true;
		for (auto v : g[u])
			if (!vis[v] && (insta[v] || !dfs2(v)))
				return false;
		vis[u] = true, insta[u] = false;
		return true;
	}
	int solve(const int r)
	{
		int ans = 0;
		dfs(0, r, 0), dfs(1, r, 0);
		for (int i = 1; i <= n; i++)
			g[i].clear();
		for (int i = 1; i <= n; i++)
		{
			if (fa[0][i] != fa[1][i])
			{
				++ans;
				if (fa[0][fa[0][i]] != fa[1][fa[0][i]])
					g[i].push_back(fa[0][i]);
				if (fa[0][fa[1][i]] != fa[1][fa[1][i]])
					g[fa[1][i]].push_back(i);
			}
			else
			{
				int tmp = i;
				while (tmp)
				{
					if (fa[0][tmp] != fa[1][tmp])
						return n + 1;
					tmp = fa[0][tmp];
				}
				tmp = i;
				while (tmp)
				{
					if (fa[0][tmp] != fa[1][tmp])
						return n + 1;
					tmp = fa[1][tmp];
				}
			}
		}
		memset(vis, 0, sizeof(bool[n + 1]));
		memset(insta, 0, sizeof(bool[n + 1]));
		for (int i = 1; i <= n; i++)
			if (fa[0][i] != fa[1][i] && !vis[i] && !dfs2(i))
				return n + 1;
		return ans;
	}
	int work()
	{
		int T;
		read(T);
		while (T--)
		{
			read(n);
			int ans = n + 1;
			memset(mp, 0, sizeof(mp));
			memset(deg, 0, sizeof(int[n + 1]));
			for (int i = 0; i < 2; i++)
				for (int j = 1; j < n; j++)
				{
					int a, b;
					read(a), read(b);
					mp[i][a][b] = mp[i][b][a] = true;
				}
			for (int i = 1; i <= n; i++)
			{
				int d = 0, t = 0;
				for (int j = 1; j <= n; j++)
					if (mp[0][i][j])
						++d, t = j;
				if (d > 1)
					continue;
				for (int j = 1; j <= n; j++)
					if (i != j)
					{
						mp[0][i][t] = mp[0][t][i] = 0;
						mp[0][i][j] = mp[0][j][i] = 1;
						ans = min(ans, solve(i) + (t != j));
						mp[0][i][j] = mp[0][j][i] = 0;
						mp[0][i][t] = mp[0][t][i] = 1;
					}
			}
			write(ans > n ? -1 : ans), putchar('\n');
		}
		return 0;
	}
}
int main()
{
	return zyt::work();
}
```