#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <numeric>

using namespace std;

const int N = 59;

bool g[2][N][N];
bool ng[2][N][N];
int leaf[N];
int pa[2][N];
int d[2][N];
int q[N];
int o[N];
int deg[N];
int color[N];

int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    int n;
    scanf("%d", &n);
    memset(g, false, sizeof g);
    for (int i = 0; i < 2; ++i) {
      for (int j = 0; j < n - 1; ++j) {
        int x, y;
        scanf("%d %d", &x, &y);
        --x;
        --y;
        g[i][x][y] = true;
        g[i][y][x] = true;
      }
    }
    for (int x = 0; x < n; ++x) {
      int cnt = 0;
      for (int y = 0; y < n; ++y) {
        if (g[0][x][y]) {
          ++cnt;
        }
      }
      leaf[x] = cnt == 1;
    }
    memcpy(ng[1], g[1], sizeof ng[1]);
    int ans = n + 1;
    for (int root = 0; root < n; ++root) {
      for (int ver = 0; ver < n; ++ver) {
        if (ver != root && !leaf[root]) {
          continue;
        }
        memcpy(ng[0], g[0], sizeof ng[0]);
        if (ver != root) {
          for (int i = 0; i < n; ++i) {
            if (ng[0][root][i]) {
              ng[0][root][i] = false;
              ng[0][i][root] = false;
            }
          }
          ng[0][root][ver] = true;
          ng[0][ver][root] = true;
        }
        for (int i = 0; i < 2; ++i) {
          pa[i][root] = root;
          d[i][root] = 0;
          q[0] = root;
          for (int qh = 0, qt = 1; qh < qt; ++qh) {
            int x = q[qh];
            for (int y = 0; y < n; ++y) {
              if (!ng[i][x][y]) {
                continue;
              }
              if (y != pa[i][x]) {
                pa[i][y] = x;
                d[i][y] = d[i][x] + 1;
                q[qt++] = y;
              }
            }
          }
        }
        iota(o, o + n, 0);
        sort(o, o + n, [](int i, int j) {
          return d[0][i] < d[0][j];
        });
        memset(color, 0, sizeof color);
        memset(deg, 0, sizeof deg);
        for (int i = 1; i < n; ++i) {
          int x = o[i];
          if (color[pa[0][x]] == 1 || pa[0][x] != pa[1][x]) {
            color[x] = 1;
          }
          ++deg[pa[0][x]];
        }
        int cur = (ver != root) + accumulate(color, color + n, 0);
        while (true) {
          bool any = false;
          for (int i = 0; i < n; ++i) {
            if (deg[i] == 0 && color[i] == 1 && color[pa[1][i]] == 0) {
              --deg[pa[0][i]];
              color[i] = 0;
              any = true;
            }
          }
          if (!any) {
            break;
          }
        }
        if (accumulate(color, color + n, 0) == 0) {
          ans = min(ans, cur);
        }
      }
    }
    if (ans == n + 1) {
      ans = -1;
    }
    printf("%d\n", ans);
  }
}

### Prompt

Please provide a cpp coded solution 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 <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <numeric>

using namespace std;

const int N = 59;

bool g[2][N][N];
bool ng[2][N][N];
int leaf[N];
int pa[2][N];
int d[2][N];
int q[N];
int o[N];
int deg[N];
int color[N];

int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    int n;
    scanf("%d", &n);
    memset(g, false, sizeof g);
    for (int i = 0; i < 2; ++i) {
      for (int j = 0; j < n - 1; ++j) {
        int x, y;
        scanf("%d %d", &x, &y);
        --x;
        --y;
        g[i][x][y] = true;
        g[i][y][x] = true;
      }
    }
    for (int x = 0; x < n; ++x) {
      int cnt = 0;
      for (int y = 0; y < n; ++y) {
        if (g[0][x][y]) {
          ++cnt;
        }
      }
      leaf[x] = cnt == 1;
    }
    memcpy(ng[1], g[1], sizeof ng[1]);
    int ans = n + 1;
    for (int root = 0; root < n; ++root) {
      for (int ver = 0; ver < n; ++ver) {
        if (ver != root && !leaf[root]) {
          continue;
        }
        memcpy(ng[0], g[0], sizeof ng[0]);
        if (ver != root) {
          for (int i = 0; i < n; ++i) {
            if (ng[0][root][i]) {
              ng[0][root][i] = false;
              ng[0][i][root] = false;
            }
          }
          ng[0][root][ver] = true;
          ng[0][ver][root] = true;
        }
        for (int i = 0; i < 2; ++i) {
          pa[i][root] = root;
          d[i][root] = 0;
          q[0] = root;
          for (int qh = 0, qt = 1; qh < qt; ++qh) {
            int x = q[qh];
            for (int y = 0; y < n; ++y) {
              if (!ng[i][x][y]) {
                continue;
              }
              if (y != pa[i][x]) {
                pa[i][y] = x;
                d[i][y] = d[i][x] + 1;
                q[qt++] = y;
              }
            }
          }
        }
        iota(o, o + n, 0);
        sort(o, o + n, [](int i, int j) {
          return d[0][i] < d[0][j];
        });
        memset(color, 0, sizeof color);
        memset(deg, 0, sizeof deg);
        for (int i = 1; i < n; ++i) {
          int x = o[i];
          if (color[pa[0][x]] == 1 || pa[0][x] != pa[1][x]) {
            color[x] = 1;
          }
          ++deg[pa[0][x]];
        }
        int cur = (ver != root) + accumulate(color, color + n, 0);
        while (true) {
          bool any = false;
          for (int i = 0; i < n; ++i) {
            if (deg[i] == 0 && color[i] == 1 && color[pa[1][i]] == 0) {
              --deg[pa[0][i]];
              color[i] = 0;
              any = true;
            }
          }
          if (!any) {
            break;
          }
        }
        if (accumulate(color, color + n, 0) == 0) {
          ans = min(ans, cur);
        }
      }
    }
    if (ans == n + 1) {
      ans = -1;
    }
    printf("%d\n", ans);
  }
}

```