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:xht37
#include <bits/stdc++.h>

#define ui unsigned int
#define ll long long
#define ul unsigned ll
#define ld long double

#define pi pair <int, int>
#define fi first
#define se second
#define mp make_pair

#define ls (p << 1)
#define rs (ls | 1)
#define md ((t[p].l + t[p].r) >> 1)

#define vi vector <int>
#define pb push_back
#define pq priority_queue

#define dbg(x) cerr << #x" = " << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)

#define fl(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)

using namespace std;

namespace io {
	const int SI = 1 << 21 | 1;
	char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
	int f, t;
	#define gc() (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin), IS == IT ? EOF : *IS++) : *IS++)
	inline void flush() {
		fwrite(OB, 1, OS - OB, stdout), OS = OB;
	}
	inline void pc(char x) {
		*OS++ = x;
		if (OS == OT) flush();
	}

	template <class I>
	inline void rd(I &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15), c = gc());
		x *= f;
	}
	template <class I>
	inline void rd(I &x, I &y) {
		rd(x), rd(y);
	}
	template <class I>
	inline void rd(I &x, I &y, I &z) {
		rd(x), rd(y), rd(z);
	}
	template <class I>
	inline void rda(I *a, int n) {
		for (int i = 1; i <= n; i++) rd(a[i]);
	}
	inline void rdc(char &c) {
		for (c = gc(); c < 33 || c > 126; c = gc());
	}
	inline void rds(char *s, int &n) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (n = 0; c >= 33 && c <= 126; s[++n] = c, c = gc());
		s[n+1] = '\0';
	}
	inline void rds(string &s) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (s.clear(); c >= 33 && c <= 126; s.pb(c), c = gc());
	}

	template <class I>
	inline void print(I x, char k = '\n') {
		if (!x) pc('0');
		if (x < 0) pc('-'), x = -x;
		while (x) ch[++t] = x % 10 + '0', x /= 10;
		while (t) pc(ch[t--]);
		pc(k);
	}
	template <class I>
	inline void print(I x, I y) {
		print(x, ' '), print(y);
	}
	template <class I>
	inline void print(I x, I y, I z) {
		print(x, ' '), print(y, ' '), print(z);
	}
	template <class I>
	inline void printa(I *a, int n) {
		for (int i = 1; i <= n; i++) print(a[i], " \n"[i==n]);
	}
	inline void printc(char c) {
		pc(c);
	}
	inline void prints(char *s, int n) {
		for (int i = 1; i <= n; i++) pc(s[i]);
		pc('\n');
	}
	inline void prints(string s) {
		int n = s.length();
		while (t < n) pc(s[t++]);
		pc('\n'), t = 0;
	}
	struct Flush {
		~Flush() {
			flush();
		}
	} flusher;
}
using io::rd;
using io::rda;
using io::rdc;
using io::rds;
using io::print;
using io::printa;
using io::printc;
using io::prints;

const int N = 57, inf = 1e9;
int n, f[2][N], d[N];
vi e[2][N], g[N];
bool v[N];
queue<int> q;

inline bool pd0() {
	for (int i = 1; i <= n; i++)
		if (e[0][i] != e[1][i]) return 0;
	return 1;
}

inline void change(int x, int y, int z) {
	e[0][x][0] = z;
	for (int &o : e[0][y])
		if (o == x) {
			swap(o, e[0][y].back());
			break;
		}
	e[0][y].pop_back(), e[0][z].pb(x);
}

void dfs(int o, int x, int fa) {
	f[o][x] = fa;
	for (int y : e[o][x])
		if (y != fa) dfs(o, y, x);
}

inline void add(int x, int y) {
	g[x].pb(y), ++d[y];
}

inline int calc(int x, int y) {
	int z = e[0][x][0], now = y != z;
	change(x, z, y);
	dfs(0, x, 0), dfs(1, x, 0);
	change(x, y, z);
	for (int i = 1; i <= n; i++)
		now += v[i] = f[0][i] != f[1][i], g[i].clear(), d[i] = 0;
	for (int i = 1; i <= n; i++)
		if (!v[i] && v[f[0][i]]) return inf;
	for (int o = 0; o < 2; o++)
		for (int i = 1; i <= n; i++)
			if (v[i] && v[f[o][i]])
				add(o ? f[o][i] : i, o ? i : f[o][i]);
	for (int i = 1; i <= n; i++)
		if (v[i] && !d[i]) q.push(i);
	while (q.size()) {
		int x = q.front();
		q.pop(), v[x] = 0;
		for (int y : g[x])
			if (!--d[y]) q.push(y);
	}
	for (int i = 1; i <= n; i++)
		if (v[i]) return inf;
	return now;
}

inline void solve() {
	rd(n);
	for (int o = 0; o < 2; o++) {
		for (int i = 1; i <= n; i++) e[o][i].clear();
		for (int i = 1, x, y; i < n; i++)
			rd(x, y), e[o][x].pb(y), e[o][y].pb(x);
		for (int i = 1; i <= n; i++)
			sort(e[o][i].begin(), e[o][i].end());
	}
	if (pd0()) return print(0);
	int ans = inf;
	for (int x = 1; x <= n; x++)
		if (e[0][x].size() == 1u)
			for (int y = 1; y <= n; y++)
				ans = min(ans, calc(x, y));
	print(ans == inf ? -1 : ans);
}

int main() {
	int T;
	rd(T);
	while (T--) solve();
	return 0;
}