#include <cstdio>
#include <algorithm>
#include <vector>

const int MN = 55;

int N, ap[MN], bp[MN], ad[MN];
std::vector<int> A[MN], B[MN];
void DFS(int u, int pa, std::vector<int> *H, int *p) {
	p[u] = pa;
	for (int v : H[u]) if (v != pa) DFS(v, u, H, p);
}

int op[MN], in[MN], que[MN], l, r;
std::vector<int> G[MN];
inline int Calc() {
	int cnt = 0;
	for (int i = 1; i <= N; ++i) cnt += op[i] = ap[i] != bp[i];
	for (int i = 1; i <= N; ++i) if (!op[i] && op[ap[i]]) return -1;
	for (int i = 1; i <= N; ++i) G[i].clear(), in[i] = 0;
	for (int i = 1; i <= N; ++i) if (op[i]) {
		if (op[ap[i]]) G[i].push_back(ap[i]), ++in[ap[i]];
		if (op[bp[i]]) G[bp[i]].push_back(i), ++in[i];
	}
	l = 1, r = 0;
	for (int i = 1; i <= N; ++i) if (op[i] && !in[i]) que[++r] = i;
	while (l <= r) {
		int u = que[l++];
		for (int v : G[u]) if (!--in[v]) que[++r] = v;
	}
	return r != cnt ? -1 : cnt;
}

int main() {
	int Tests;
	scanf("%d", &Tests);
	while (Tests--) {
		scanf("%d", &N);
		int Ans = N + 1;
		for (int i = 1; i <= N; ++i) A[i].clear(), B[i].clear(), ad[i] = 0;
		for (int i = 1, x, y; i < N; ++i)
			scanf("%d%d", &x, &y), A[x].push_back(y), A[y].push_back(x), ++ad[x], ++ad[y];
		for (int i = 1, x, y; i < N; ++i)
			scanf("%d%d", &x, &y), B[x].push_back(y), B[y].push_back(x);
		DFS(1, 0, A, ap), DFS(1, 0, B, bp);
		int ok = 1;
		for (int i = 1; i <= N; ++i) if (ap[i] != bp[i]) ok = 0;
		if (ok) { puts("0"); continue; }
		for (int i = 1; i <= N; ++i) if (ad[i] == 1) {
			DFS(i, 0, B, bp);
			for (int j = 1; j <= N; ++j) if (j != i) {
				DFS(j, 0, A, ap), ap[j] = i, ap[i] = 0;
				int ans = Calc();
				if (~ans) Ans = std::min(Ans, ans + 1);
			}
		}
		printf("%d\n", Ans > N ? -1 : Ans);
	}
	return 0;
}
### Prompt

Please create a solution in Cpp 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 <cstdio>
#include <algorithm>
#include <vector>

const int MN = 55;

int N, ap[MN], bp[MN], ad[MN];
std::vector<int> A[MN], B[MN];
void DFS(int u, int pa, std::vector<int> *H, int *p) {
	p[u] = pa;
	for (int v : H[u]) if (v != pa) DFS(v, u, H, p);
}

int op[MN], in[MN], que[MN], l, r;
std::vector<int> G[MN];
inline int Calc() {
	int cnt = 0;
	for (int i = 1; i <= N; ++i) cnt += op[i] = ap[i] != bp[i];
	for (int i = 1; i <= N; ++i) if (!op[i] && op[ap[i]]) return -1;
	for (int i = 1; i <= N; ++i) G[i].clear(), in[i] = 0;
	for (int i = 1; i <= N; ++i) if (op[i]) {
		if (op[ap[i]]) G[i].push_back(ap[i]), ++in[ap[i]];
		if (op[bp[i]]) G[bp[i]].push_back(i), ++in[i];
	}
	l = 1, r = 0;
	for (int i = 1; i <= N; ++i) if (op[i] && !in[i]) que[++r] = i;
	while (l <= r) {
		int u = que[l++];
		for (int v : G[u]) if (!--in[v]) que[++r] = v;
	}
	return r != cnt ? -1 : cnt;
}

int main() {
	int Tests;
	scanf("%d", &Tests);
	while (Tests--) {
		scanf("%d", &N);
		int Ans = N + 1;
		for (int i = 1; i <= N; ++i) A[i].clear(), B[i].clear(), ad[i] = 0;
		for (int i = 1, x, y; i < N; ++i)
			scanf("%d%d", &x, &y), A[x].push_back(y), A[y].push_back(x), ++ad[x], ++ad[y];
		for (int i = 1, x, y; i < N; ++i)
			scanf("%d%d", &x, &y), B[x].push_back(y), B[y].push_back(x);
		DFS(1, 0, A, ap), DFS(1, 0, B, bp);
		int ok = 1;
		for (int i = 1; i <= N; ++i) if (ap[i] != bp[i]) ok = 0;
		if (ok) { puts("0"); continue; }
		for (int i = 1; i <= N; ++i) if (ad[i] == 1) {
			DFS(i, 0, B, bp);
			for (int j = 1; j <= N; ++j) if (j != i) {
				DFS(j, 0, A, ap), ap[j] = i, ap[i] = 0;
				int ans = Calc();
				if (~ans) Ans = std::min(Ans, ans + 1);
			}
		}
		printf("%d\n", Ans > N ? -1 : Ans);
	}
	return 0;
}
```