#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for (int i = (a); i <= (b); ++ i)
#define rrp(i,a,b) for (int i = (a); i >= (b); -- i)
#define gc() getchar()
#define fir first
#define sec second
typedef pair<int,int> pii;
typedef double db;
typedef long long ll;
template <typename tp>
inline void read(tp& x) {
  x = 0; char tmp; bool key = 0;
  for (tmp = gc(); !isdigit(tmp); tmp = gc())
    key = (tmp == '-');
  for (; isdigit(tmp); tmp = gc()) 
    x = (x << 3) + (x << 1) + (tmp ^ '0');
  if (key) x = -x;
}
template <typename tp>
inline void ckmn(tp& x,tp y) {
  x = x < y ? x : y;
}
template <typename tp>
inline void ckmx(tp& x,tp y) {
  x = x < y ? y : x;
}

const int N = 55;
int n;
struct edge {
  int la,b;
};
struct graph {
  edge con[N << 1];
  int tot, fir[N];
  void add_edge(int from,int to) {
    con[++tot] = (edge) {fir[from], to};
    fir[from] = tot;
  }
  void dfs(int pos,int *fa) {
    for (int i = fir[pos]; i; i = con[i].la) {
      if (con[i].b == fa[pos]) continue;
      fa[con[i].b] = pos;
      dfs(con[i].b, fa);
    }
  }
  void init() {
    memset(fir, 0, sizeof fir);
    tot = 0;
  }
  int deg[N];
  bool bfs() {
    rep (i, 1, n) deg[i] = 0;
    rep (pos, 1, n)
      for (int i = fir[pos]; i; i = con[i].la)
	++ deg[con[i].b];
    queue<int> q;
    rep (i, 1, n) if (!deg[i])
      q.push(i);
    int cnt = 0;
    for (int pos; !q.empty(); q.pop()) {
      pos = q.front();
      ++ cnt;
      for (int i = fir[pos]; i; i = con[i].la) {
	if (-- deg[con[i].b] == 0)
	  q.push(con[i].b);
      }
    }
    return cnt == n;
  }
} t1, t2, g;
int ans;
int fa[2][N], vis[N];
void doit0(int rt) {
  memset(fa, 0, sizeof fa);
  t1.dfs(rt, fa[0]);
  t2.dfs(rt, fa[1]);
  rep (i, 1, n) vis[i] = 0;
  rep (i, 1, n) {
    if (fa[0][i] == fa[1][i]) {
      vis[i] = 1;
      for (int j = fa[0][i]; j; j = fa[0][j]) {
	if (fa[0][j] != fa[1][j]) return;
	vis[j] = 1;
      }
    }
  }
  g.init();
  rep (i, 1, n) if (!vis[i]) {
    if (!vis[fa[0][i]]) g.add_edge(fa[0][i], i);
    if (!vis[fa[1][i]]) g.add_edge(i, fa[1][i]);
  }
  if (!g.bfs()) return;
  int num = n;
  rep (i, 1, n) if (vis[i]) num --;
  ans = min(ans, num);
}
void doit1(int rt1,int rt2) {
  memset(fa, 0, sizeof fa);
  t1.dfs(rt1, fa[0]);
  t2.dfs(rt2, fa[1]);
  g.init();
  rep (i, 1, n) {
    if (fa[0][i]) g.add_edge(fa[0][i], i);
    if (fa[1][i]) g.add_edge(i, fa[1][i]);
  }
  if (!g.bfs()) return;
  ans = min(ans, n);
}
void solve() {
  int x,y;
  read(n);
  t1.init(), t2.init();
  rep (i, 1, n-1) {
    read(x), read(y);
    t1.add_edge(x, y);
    t1.add_edge(y, x);
  }
  rep (i, 1, n-1) {
    read(x), read(y);
    t2.add_edge(x, y);
    t2.add_edge(y, x);    
  }
  ans = n+1;
  rep (i, 1, n)
    doit0(i);
  rep (i, 1, n) rep (j, 1, n)
    doit1(i, j);
  if (ans > n) puts("-1");
  else printf("%d\n", ans);
}
int main() {
  int T;
  read(T);
  while (T --)
    solve();
  return 0;
}

### Prompt

In CPP, your task is to solve 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 <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for (int i = (a); i <= (b); ++ i)
#define rrp(i,a,b) for (int i = (a); i >= (b); -- i)
#define gc() getchar()
#define fir first
#define sec second
typedef pair<int,int> pii;
typedef double db;
typedef long long ll;
template <typename tp>
inline void read(tp& x) {
  x = 0; char tmp; bool key = 0;
  for (tmp = gc(); !isdigit(tmp); tmp = gc())
    key = (tmp == '-');
  for (; isdigit(tmp); tmp = gc()) 
    x = (x << 3) + (x << 1) + (tmp ^ '0');
  if (key) x = -x;
}
template <typename tp>
inline void ckmn(tp& x,tp y) {
  x = x < y ? x : y;
}
template <typename tp>
inline void ckmx(tp& x,tp y) {
  x = x < y ? y : x;
}

const int N = 55;
int n;
struct edge {
  int la,b;
};
struct graph {
  edge con[N << 1];
  int tot, fir[N];
  void add_edge(int from,int to) {
    con[++tot] = (edge) {fir[from], to};
    fir[from] = tot;
  }
  void dfs(int pos,int *fa) {
    for (int i = fir[pos]; i; i = con[i].la) {
      if (con[i].b == fa[pos]) continue;
      fa[con[i].b] = pos;
      dfs(con[i].b, fa);
    }
  }
  void init() {
    memset(fir, 0, sizeof fir);
    tot = 0;
  }
  int deg[N];
  bool bfs() {
    rep (i, 1, n) deg[i] = 0;
    rep (pos, 1, n)
      for (int i = fir[pos]; i; i = con[i].la)
	++ deg[con[i].b];
    queue<int> q;
    rep (i, 1, n) if (!deg[i])
      q.push(i);
    int cnt = 0;
    for (int pos; !q.empty(); q.pop()) {
      pos = q.front();
      ++ cnt;
      for (int i = fir[pos]; i; i = con[i].la) {
	if (-- deg[con[i].b] == 0)
	  q.push(con[i].b);
      }
    }
    return cnt == n;
  }
} t1, t2, g;
int ans;
int fa[2][N], vis[N];
void doit0(int rt) {
  memset(fa, 0, sizeof fa);
  t1.dfs(rt, fa[0]);
  t2.dfs(rt, fa[1]);
  rep (i, 1, n) vis[i] = 0;
  rep (i, 1, n) {
    if (fa[0][i] == fa[1][i]) {
      vis[i] = 1;
      for (int j = fa[0][i]; j; j = fa[0][j]) {
	if (fa[0][j] != fa[1][j]) return;
	vis[j] = 1;
      }
    }
  }
  g.init();
  rep (i, 1, n) if (!vis[i]) {
    if (!vis[fa[0][i]]) g.add_edge(fa[0][i], i);
    if (!vis[fa[1][i]]) g.add_edge(i, fa[1][i]);
  }
  if (!g.bfs()) return;
  int num = n;
  rep (i, 1, n) if (vis[i]) num --;
  ans = min(ans, num);
}
void doit1(int rt1,int rt2) {
  memset(fa, 0, sizeof fa);
  t1.dfs(rt1, fa[0]);
  t2.dfs(rt2, fa[1]);
  g.init();
  rep (i, 1, n) {
    if (fa[0][i]) g.add_edge(fa[0][i], i);
    if (fa[1][i]) g.add_edge(i, fa[1][i]);
  }
  if (!g.bfs()) return;
  ans = min(ans, n);
}
void solve() {
  int x,y;
  read(n);
  t1.init(), t2.init();
  rep (i, 1, n-1) {
    read(x), read(y);
    t1.add_edge(x, y);
    t1.add_edge(y, x);
  }
  rep (i, 1, n-1) {
    read(x), read(y);
    t2.add_edge(x, y);
    t2.add_edge(y, x);    
  }
  ans = n+1;
  rep (i, 1, n)
    doit0(i);
  rep (i, 1, n) rep (j, 1, n)
    doit1(i, j);
  if (ans > n) puts("-1");
  else printf("%d\n", ans);
}
int main() {
  int T;
  read(T);
  while (T --)
    solve();
  return 0;
}

```