#include <bits/stdc++.h>

using namespace std;

const int maxn = 55;

int T, n;
int E1[maxn][maxn], E2[maxn][maxn], E[maxn][maxn], deg[maxn];
int tag[maxn], _deg[maxn], _E1[maxn][maxn], _E2[maxn][maxn];

void dfs_tag(int u, int f) {
	tag[u] = 1;
	for (int v = 1; v <= n; v++) {
		if (E1[u][v] && E2[u][v] && v != f) {
			dfs_tag(v, u);
		}
	}
}

void dfs1(int u, int f) {
	for (int v = 1; v <= n; v++) {
		if (E1[u][v] && v != f) {
			if (!tag[u]) {
				++ E[u][v];
				++ deg[v];
			}
			dfs1(v, u);
		}
	}
}

void dfs2(int u, int f) {
	for (int v = 1; v <= n; v++) {
		if (E2[u][v] && v != f) {
			if (!tag[u]) {
				++ E[v][u];
				++ deg[u];
			}
			dfs2(v, u);
		}
	}
}

int main() {
	scanf("%d", &T);
	while (T --) {
		int ans = 0x3f3f3f3f;
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				_E1[i][j] = _E2[i][j] = 0;
			}
			_deg[i] = 0;
		}
		for (int i = 1; i < n; i++) {
			int u, v; scanf("%d%d", &u, &v);
			_E1[u][v] = _E1[v][u] = 1;
			++ _deg[u], ++ _deg[v];
		}
		for (int i = 1; i < n; i++) {
			int u, v; scanf("%d%d", &u, &v);
			_E2[u][v] = _E2[v][u] = 1;
		}
		int dif = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				if (_E1[i][j] != _E2[i][j]) {
					dif = 1;
				}
			}
		}
		if (!dif) {
			puts("0");
			continue;
		}
		for (int s = 1; s <= n; s++) {
			if (_deg[s] != 1) continue;
			int f = 0;
			for (int i = 1; i <= n; i++) if (_E1[s][i]) f = i;
			for (int t = 1; t <= n; t++) {
				if (t == s) continue;
				for (int i = 1; i <= n; i++) {
					for (int j = 1; j <= n; j++) {
						E[i][j] = 0;
						E1[i][j] = _E1[i][j];
						E2[i][j] = _E2[i][j]; 
					}
					deg[i] = 0;
				}
				E1[s][f] = E1[f][s] = 0, E1[s][t] = E1[t][s] = 1;
				int cnt = 0;
				for (int i = 1; i <= n; i++) tag[i] = 0;
				dfs_tag(s, 0);
				for (int i = 1; i <= n; i++) if (tag[i]) ++ cnt;
				dfs1(s, 0), dfs2(s, 0);
				queue<int> Q;
				for (int i = 1; i <= n; i++) {
					if (!tag[i] && !deg[i]) Q.push(i);
				}
				while (!Q.empty()) {
					int u = Q.front(); Q.pop();
					for (int v = 1; v <= n; v++) {
						if (E[u][v]) {
							deg[v] -= E[u][v];
							if (!deg[v]) Q.push(v);
						}
					}
				}
				int ok = 1;
				for (int i = 1; i <= n; i++) if (!tag[i] && deg[i]) ok = 0;
				if (ok) ans = min(ans, n - cnt);
			}
		}
		if (ans < 0x3f3f3f3f) printf("%d\n", ans + 1);
		else puts("-1");
	}
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <bits/stdc++.h>

using namespace std;

const int maxn = 55;

int T, n;
int E1[maxn][maxn], E2[maxn][maxn], E[maxn][maxn], deg[maxn];
int tag[maxn], _deg[maxn], _E1[maxn][maxn], _E2[maxn][maxn];

void dfs_tag(int u, int f) {
	tag[u] = 1;
	for (int v = 1; v <= n; v++) {
		if (E1[u][v] && E2[u][v] && v != f) {
			dfs_tag(v, u);
		}
	}
}

void dfs1(int u, int f) {
	for (int v = 1; v <= n; v++) {
		if (E1[u][v] && v != f) {
			if (!tag[u]) {
				++ E[u][v];
				++ deg[v];
			}
			dfs1(v, u);
		}
	}
}

void dfs2(int u, int f) {
	for (int v = 1; v <= n; v++) {
		if (E2[u][v] && v != f) {
			if (!tag[u]) {
				++ E[v][u];
				++ deg[u];
			}
			dfs2(v, u);
		}
	}
}

int main() {
	scanf("%d", &T);
	while (T --) {
		int ans = 0x3f3f3f3f;
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				_E1[i][j] = _E2[i][j] = 0;
			}
			_deg[i] = 0;
		}
		for (int i = 1; i < n; i++) {
			int u, v; scanf("%d%d", &u, &v);
			_E1[u][v] = _E1[v][u] = 1;
			++ _deg[u], ++ _deg[v];
		}
		for (int i = 1; i < n; i++) {
			int u, v; scanf("%d%d", &u, &v);
			_E2[u][v] = _E2[v][u] = 1;
		}
		int dif = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				if (_E1[i][j] != _E2[i][j]) {
					dif = 1;
				}
			}
		}
		if (!dif) {
			puts("0");
			continue;
		}
		for (int s = 1; s <= n; s++) {
			if (_deg[s] != 1) continue;
			int f = 0;
			for (int i = 1; i <= n; i++) if (_E1[s][i]) f = i;
			for (int t = 1; t <= n; t++) {
				if (t == s) continue;
				for (int i = 1; i <= n; i++) {
					for (int j = 1; j <= n; j++) {
						E[i][j] = 0;
						E1[i][j] = _E1[i][j];
						E2[i][j] = _E2[i][j]; 
					}
					deg[i] = 0;
				}
				E1[s][f] = E1[f][s] = 0, E1[s][t] = E1[t][s] = 1;
				int cnt = 0;
				for (int i = 1; i <= n; i++) tag[i] = 0;
				dfs_tag(s, 0);
				for (int i = 1; i <= n; i++) if (tag[i]) ++ cnt;
				dfs1(s, 0), dfs2(s, 0);
				queue<int> Q;
				for (int i = 1; i <= n; i++) {
					if (!tag[i] && !deg[i]) Q.push(i);
				}
				while (!Q.empty()) {
					int u = Q.front(); Q.pop();
					for (int v = 1; v <= n; v++) {
						if (E[u][v]) {
							deg[v] -= E[u][v];
							if (!deg[v]) Q.push(v);
						}
					}
				}
				int ok = 1;
				for (int i = 1; i <= n; i++) if (!tag[i] && deg[i]) ok = 0;
				if (ok) ans = min(ans, n - cnt);
			}
		}
		if (ans < 0x3f3f3f3f) printf("%d\n", ans + 1);
		else puts("-1");
	}
	return 0;
}
```