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;
const int MAXN = 105;
typedef long long ll;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
int n, fa[MAXN], fb[MAXN];
vector <int> a[MAXN], b[MAXN];
void dfsa(int pos, int f) {
	fa[pos] = f;
	for (auto x : a[pos])
		if (x != f) dfsa(x, pos);
}
void dfsb(int pos, int f) {
	fb[pos] = f;
	for (auto x : b[pos])
		if (x != f) dfsb(x, pos);
}
bool vis[MAXN], unlock[MAXN], used[MAXN];
void dfs(int pos) {
	vis[pos] = true;
	for (auto x : a[pos])
		if (fa[x] == pos && fb[x] == pos) dfs(x);
}
bool root(int pos) {
	memset(unlock, false, sizeof(unlock)), unlock[pos] = true;
	memset(used, false, sizeof(used));
	for (int t = 1; t <= n; t++) {
		int pos = 0;
		for (int i = 1; i <= n; i++)
			if (!used[i] && unlock[i]) {
				int d = 0;
				for (auto x : a[i])
					if (!used[x]) d++;
				if (d <= 1) pos = i;
			}
		if (pos == 0) return false;
		used[pos] = true;
		for (auto x : b[pos])
			unlock[x] = true;
	}
	return true;
}
int work(int pos) {
	memset(vis, false, sizeof(vis)), dfs(pos);
	memcpy(unlock, vis, sizeof(vis));
	for (int i = 1; i <= n; i++)
		if (vis[i]) {
			for (auto x : b[i])
				unlock[x] = true;
		}
	memset(used, false, sizeof(used));
	int cnt = 0;
	for (int i = 1; i <= n; i++)
		if (!vis[i]) cnt++;
	for (int t = 1; t <= cnt; t++) {
		int pos = 0;
		for (int i = 1; i <= n; i++)
			if (!vis[i] && !used[i] && unlock[i]) {
				int d = 0;
				for (auto x : a[i])
					if (!used[x]) d++;
				if (d <= 1) pos = i;
			}
		if (pos == 0) return n + 1;
		used[pos] = true;
		for (auto x : b[pos])
			unlock[x] = true;
	}
	return cnt;
}
void solve() {
	int ans = n + 1;
	for (int i = 1; i <= n; i++) {
		dfsa(i, 0);
		dfsb(i, 0);
		chkmin(ans, work(i));
		if (root(i)) chkmin(ans, n);
	}
	if (ans == n + 1) puts("-1");
	else cout << ans << endl;
}
int main() {
	int T; read(T);
	while (T--) {
		read(n);
		for (int i = 1; i <= n; i++) {
			a[i].clear();
			b[i].clear();
		}
		for (int i = 1; i <= n - 1; i++) {
			int x, y; read(x), read(y);
			a[x].push_back(y);
			a[y].push_back(x);
		}
		for (int i = 1; i <= n - 1; i++) {
			int x, y; read(x), read(y);
			b[x].push_back(y);
			b[y].push_back(x);
		}
		solve();
	}
	return 0;
}