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:
// ====================================
//   author: M_sea
//   website: https://m-sea-blog.com/
// ====================================
#include <bits/stdc++.h>
#define file(x) freopen(#x".in","r",stdin); freopen(#x".out","w",stdout)
#define debug(...) fprintf(stderr,__VA_ARGS__)
using namespace std;
typedef long long ll;

int read() {
	int X=0,w=1; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
	while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
	return X*w;
}

const int N=50+10;
const int inf=0x3f3f3f3f;

int n;

struct V {
	vector<int> E[N];
	int deg[N];
	void addEdge(int u,int v) {
		E[u].emplace_back(v),++deg[v];
	}

	void init() {
		for (int i=1;i<=n;++i) E[i].clear(),deg[i]=0;
	}

	int fa[N];
	void dfs(int u,int f) {
		fa[u]=f;
		for (int v:E[u]) if (v!=f) dfs(v,u);
	}
} A,B,G;

int check(int u) { return A.fa[u]==B.fa[u]; }
int calc() {
	int res=0;
	for (int i=1;i<=n;++i) if (!check(i)) ++res;
	G.init();
	for (int i=1;i<=n;++i) {
		if (check(i)) {
			if (!check(A.fa[i])) return inf;
			else continue;
		}
		if (!check(A.fa[i])) G.addEdge(i,A.fa[i]);
		if (!check(B.fa[i])) G.addEdge(B.fa[i],i);
	}
	queue<int> Q; int cnt=0;
	for (int i=1;i<=n;++i) if (!G.deg[i]) Q.push(i);
	while (!Q.empty()) {
		int u=Q.front(); Q.pop(); ++cnt;
		for (int v:G.E[u])
			if (!--G.deg[v]) Q.push(v);
	}
	return cnt==n?res:inf;
}

int main() {
	int T=read();
	while (T--) {
		n=read(); A.init(),B.init();
		for (int i=1;i<n;++i) {
			int u=read(),v=read();
			A.addEdge(u,v),A.addEdge(v,u);
		}
		for (int i=1;i<n;++i) {
			int u=read(),v=read();
			B.addEdge(u,v),B.addEdge(v,u);
		}
		int ans=inf;
		for (int i=1;i<=n;++i) {
			A.dfs(i,0),B.dfs(i,0),ans=min(ans,calc());
			if (A.deg[i]!=1) continue;
			for (int j=1;j<=n;++j) {
				if (i==j) continue;
				A.dfs(j,i),A.fa[i]=0,ans=min(ans,calc()+1);
			}
		}
		if (ans==inf) puts("-1");
		else printf("%d\n",ans);
	}
	return 0;
}