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;

int EA[53][53] , EB[53][53] , faA[53] , faB[53] , T , N; bool same[53];

void find_same(int x , int p){
	same[x] = 1;
	for(int i = 1 ; i <= N ; ++i)
		if(EA[x][i] && EB[x][i] && i != p) find_same(i , x);
}

void getfa(int x , int p , int *fa , int E[][53]){
	fa[x] = p;
	for(int i = 1 ; i <= N ; ++i)
		if(E[x][i] && i != p) getfa(i , x , fa , E);
}

vector < int > nxt[53]; int du[53];

int main(){
	for(cin >> T ; T ; --T){
		cin >> N; int ans = 200;
		memset(EA , 0 , sizeof(EA));
		for(int i = 1 ; i < N ; ++i){int x , y; cin >> x >> y; EA[x][y] = EA[y][x] = 1;}
		memset(EB , 0 , sizeof(EB));
		for(int i = 1 ; i < N ; ++i){int x , y; cin >> x >> y; EB[x][y] = EB[y][x] = 1;}
		for(int i = 1 ; i <= N ; ++i){
			int cnt = 0;
			for(int j = 1 ; j <= N ; ++j) cnt += EA[i][j];
			if(cnt != 1) continue;
			int tar; for(int j = 1 ; j <= N ; ++j) if(EA[i][j]) tar = j;
			EA[i][tar] = EA[tar][i] = 0;
			for(int j = 1 ; j <= N ; ++j)
				if(j != i){
					EA[i][j] = EA[j][i] = 1; memset(same , 0 , sizeof(same));
					find_same(i , 0); memset(du , 0 , sizeof(du));
					getfa(i , 0 , faA , EA); getfa(i , 0 , faB , EB);
					int num = 0; for(int i = 1 ; i <= N ; ++i) nxt[i].clear();
					for(int i = 1 ; i <= N ; ++i)
						if(!same[i]){
							++num;
							if(!same[faA[i]]){nxt[i].push_back(faA[i]); ++du[faA[i]];}
							if(!same[faB[i]]){nxt[faB[i]].push_back(i); ++du[i];}
						}
					queue < int > q; int num1 = 0;
					for(int i = 1 ; i <= N ; ++i) if(!same[i] && !du[i]) q.push(i);
					while(!q.empty()){
						int t = q.front(); q.pop(); ++num1;
						for(auto p : nxt[t]) if(!--du[p]) q.push(p);
					}
					if(num1 == num) ans = min(ans , num + (j != tar));
					EA[i][j] = EA[j][i] = 0;
				}
			EA[i][tar] = EA[tar][i] = 1;
		}
		cout << (ans == 200 ? -1 : ans) << endl;
	}
	return 0;
}