#include <bits/stdc++.h>

using namespace std;

const int MX = 55;

template <typename T> void cmin(T &x, const T &y)
{
	if(y < x) x = y;
}

template <typename T> void cmax(T &x, const T &y)
{
	if(y > x) x = y;
}

template <typename T> void read(T &x)
{
	x = 0; char c = getchar(); bool f = 0;
	while(!isdigit(c) && c!='-') c = getchar();
	if(c == '-') f = 1, c = getchar();
	while(isdigit(c)) x = x*10+c-'0', c = getchar();
	if(f) x = -x;
}

struct GRAPH
{
	vector<int> v[MX];
	bool e[MX][MX];
	int deg[MX];

	void deleg(int a, int b)
	{
		for(auto i=v[a].begin(); i!=v[a].end(); i++)
			if(*i == b)
			{
				v[a].erase(i);
				break;
			}
		e[a][b] = 0;
		deg[b]--;
	}

	void addeg(int a, int b)
	{
		v[a].push_back(b);
		e[a][b] = 1;
		deg[b]++;
	}

	void init()
	{
		memset(e, 0, sizeof(e));
		memset(deg, 0, sizeof(deg));
		for(int i=0; i<MX; i++) v[i].clear();
	}

	bool no_loop(int n)
	{
		int cnt = 0;
		vector<int> stk;
		for(int i=1; i<=n; i++)
			if(!deg[i])
				stk.push_back(i);
		while(!stk.empty())
		{
			int x = stk.back();
			cnt++;
			stk.pop_back();
			for(auto y : v[x])
				if(!(--deg[y]))
					stk.push_back(y);
		}
		return (cnt==n);
	}
} A, B, C;

int n;

void input()
{
	A.init();
	B.init();
	read(n);
	for(int i=1; i<n; i++)
	{
		int a, b;
		read(a), read(b);
		A.addeg(a, b);
		A.addeg(b, a);
	}
	for(int i=1; i<n; i++)
	{
		int a, b;
		read(a), read(b);
		B.addeg(a, b);
		B.addeg(b, a);
	}
}

bool fxd[MX];
int faa[MX], fab[MX];

void getfa(int x, int f, const GRAPH &g, int *fa)
{
	fa[x] = f;
	for(auto y : g.v[x])
		if(y != f)
			getfa(y, x, g, fa);
}

void dfs(int x, int f)
{
	fxd[x] = 1;
	for(auto y : A.v[x])
	{
		if(y == f) continue;
		if(B.e[x][y]) dfs(y, x);
	}
}

int calc(int root)
{
	memset(fxd, 0, sizeof(fxd));
	getfa(root, 0, A, faa);
	getfa(root, 0, B, fab);
	dfs(root, 0);
	C.init();
	int cnt = 0;
	for(int i=1; i<=n; i++)
	{
		if(!fxd[i]) cnt++;
		if(!fxd[i] && !fxd[faa[i]]) C.addeg(i, faa[i]);
		if(!fxd[i] && !fxd[fab[i]]) C.addeg(fab[i], i);
	}
	if(C.no_loop(n)) return cnt;
	else return n+1;
}

void work()
{
	int ans = n+1;
	for(int i=1; i<=n; i++) cmin(ans, calc(i));
	for(int i=1; i<=n; i++)
		if(A.deg[i] == 1)
			for(int j=1; j<=n; j++)
				if(i != j)
				{
					int f = *A.v[i].begin();
					if(f == j) continue;
					A.deleg(i, f);
					A.deleg(f, i);
					A.addeg(i, j);
					A.addeg(j, i);
					cmin(ans, calc(i)+1);
					A.deleg(i, j);
					A.deleg(j, i);
					A.addeg(i, f);
					A.addeg(f, i);
				}
	if(ans == n+1) puts("-1");
	else printf("%d\n", ans);
}

int main()
{
	int t;
	read(t);
	while(t--)
	{
		input();
		work();
	}
	return 0;
}

### 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 <bits/stdc++.h>

using namespace std;

const int MX = 55;

template <typename T> void cmin(T &x, const T &y)
{
	if(y < x) x = y;
}

template <typename T> void cmax(T &x, const T &y)
{
	if(y > x) x = y;
}

template <typename T> void read(T &x)
{
	x = 0; char c = getchar(); bool f = 0;
	while(!isdigit(c) && c!='-') c = getchar();
	if(c == '-') f = 1, c = getchar();
	while(isdigit(c)) x = x*10+c-'0', c = getchar();
	if(f) x = -x;
}

struct GRAPH
{
	vector<int> v[MX];
	bool e[MX][MX];
	int deg[MX];

	void deleg(int a, int b)
	{
		for(auto i=v[a].begin(); i!=v[a].end(); i++)
			if(*i == b)
			{
				v[a].erase(i);
				break;
			}
		e[a][b] = 0;
		deg[b]--;
	}

	void addeg(int a, int b)
	{
		v[a].push_back(b);
		e[a][b] = 1;
		deg[b]++;
	}

	void init()
	{
		memset(e, 0, sizeof(e));
		memset(deg, 0, sizeof(deg));
		for(int i=0; i<MX; i++) v[i].clear();
	}

	bool no_loop(int n)
	{
		int cnt = 0;
		vector<int> stk;
		for(int i=1; i<=n; i++)
			if(!deg[i])
				stk.push_back(i);
		while(!stk.empty())
		{
			int x = stk.back();
			cnt++;
			stk.pop_back();
			for(auto y : v[x])
				if(!(--deg[y]))
					stk.push_back(y);
		}
		return (cnt==n);
	}
} A, B, C;

int n;

void input()
{
	A.init();
	B.init();
	read(n);
	for(int i=1; i<n; i++)
	{
		int a, b;
		read(a), read(b);
		A.addeg(a, b);
		A.addeg(b, a);
	}
	for(int i=1; i<n; i++)
	{
		int a, b;
		read(a), read(b);
		B.addeg(a, b);
		B.addeg(b, a);
	}
}

bool fxd[MX];
int faa[MX], fab[MX];

void getfa(int x, int f, const GRAPH &g, int *fa)
{
	fa[x] = f;
	for(auto y : g.v[x])
		if(y != f)
			getfa(y, x, g, fa);
}

void dfs(int x, int f)
{
	fxd[x] = 1;
	for(auto y : A.v[x])
	{
		if(y == f) continue;
		if(B.e[x][y]) dfs(y, x);
	}
}

int calc(int root)
{
	memset(fxd, 0, sizeof(fxd));
	getfa(root, 0, A, faa);
	getfa(root, 0, B, fab);
	dfs(root, 0);
	C.init();
	int cnt = 0;
	for(int i=1; i<=n; i++)
	{
		if(!fxd[i]) cnt++;
		if(!fxd[i] && !fxd[faa[i]]) C.addeg(i, faa[i]);
		if(!fxd[i] && !fxd[fab[i]]) C.addeg(fab[i], i);
	}
	if(C.no_loop(n)) return cnt;
	else return n+1;
}

void work()
{
	int ans = n+1;
	for(int i=1; i<=n; i++) cmin(ans, calc(i));
	for(int i=1; i<=n; i++)
		if(A.deg[i] == 1)
			for(int j=1; j<=n; j++)
				if(i != j)
				{
					int f = *A.v[i].begin();
					if(f == j) continue;
					A.deleg(i, f);
					A.deleg(f, i);
					A.addeg(i, j);
					A.addeg(j, i);
					cmin(ans, calc(i)+1);
					A.deleg(i, j);
					A.deleg(j, i);
					A.addeg(i, f);
					A.addeg(f, i);
				}
	if(ans == n+1) puts("-1");
	else printf("%d\n", ans);
}

int main()
{
	int t;
	read(t);
	while(t--)
	{
		input();
		work();
	}
	return 0;
}

```