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 N = 52;
int n;
vector<int> G1[N], G2[N], T[N];
int fix[N], fa[N], vis[N], ins[N];
void dfs1(int x, int pre, int ign) {
  fa[x] = pre;
  for (auto y : G1[x]) if (y != pre && y != ign) {
    dfs1(y, x, ign);
  }
}
void dfs2(int x, int pre) {
  if (fa[x] != pre) return;
  fix[x] = 1;
  for (auto y : G2[x]) if (y != pre) {
    dfs2(y, x);
  }
}
void dfs3(int x, int pre, int ign) {
  if (!fix[x] && !fix[pre]) T[x].push_back(pre);
  for (auto y : G1[x]) if (y != pre && y != ign) {
    dfs3(y, x, ign);
  }
}
void dfs4(int x, int pre) {
  if (!fix[x] && !fix[pre]) T[pre].push_back(x);
  for (auto y : G2[x]) if (y != pre) {
    dfs4(y, x);
  }
}
int dfs5(int x) {
  vis[x] = ins[x] = 1;
  for (auto y : T[x]) {
    if (ins[y]) {
      return 1;
    } else if (!vis[y] && dfs5(y)) {
      return 1;
    }
  }
  ins[x] = 0;
  return 0;
}
inline int solve(int rt, int rt2) {
  memset(fix, 0, sizeof fix);
  memset(fa, 0, sizeof fa);
  memset(vis, 0, sizeof vis);
  memset(ins, 0, sizeof ins);
  for (int i = 1; i <= n; i++) {
    T[i].clear();
  }
  dfs1(rt2, rt, rt);
  dfs2(rt, 0);
  dfs3(rt2, rt, rt);
  dfs4(rt, 0);
  for (int i = 1; i <= n; i++) {
    if (!vis[i] && dfs5(i)) return -1;
  }
  return n - accumulate(fix + 1, fix + 1 + n, 0);
}
inline void Main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    G1[i].clear();
    G2[i].clear();
  }
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    G1[x].push_back(y);
    G1[y].push_back(x);
  }
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    G2[x].push_back(y);
    G2[y].push_back(x);
  }
  int ans = 999;
  for (int i = 1; i <= n; i++) if (G1[i].size() == 1) {
    for (int j = 1; j <= n; j++) if (j != i) {
      int x = solve(i, j);
      if (~x) ans = min(ans, x + (j != G1[i][0]));
    }
  }
  printf("%d\n", ans == 999 ? -1 : ans);
}
int main() {
  int Test;
  scanf("%d", &Test);
  while (Test--) {
    Main();
  }
}