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 <cstring>
#include <cctype>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 5.5e1+5;
int n, pa[N], pb[N], d[N], ans; vector<int> ea[N], eb[N];
void dfs(int x, int p, vector<int> *e, int *fa)
{ fa[x] = p; for(auto y:e[x]) if(y != p) dfs(y, x, e, fa); }
int solve()
{
    static int op[N], in[N], q[N]; static vector<int> e[N]; int ret = 0, l = 1, r = 0;
    for(int i = 1; i <= n; ++i) ret += op[i] = pa[i] != pb[i], in[i] = 0, e[i].clear();
    for(int i = 1; i <= n; ++i) if(!op[i]&&op[pa[i]]) return -1;
    for(int i = 1; i <= n; ++i) if(op[i])
    {
        if(op[pa[i]]) e[i].push_back(pa[i]), ++in[pa[i]];
        if(op[pb[i]]) e[pb[i]].push_back(i), ++in[i];
    }
    for(int i = 1; i <= n; ++i) if(op[i]&&!in[i]) q[++r] = i;
    while(l <= r)
    {
        int x = q[l++];
        for(auto y:e[x]) if(!--in[y]) q[++r] = y;
    }
    return r^ret?-1:ret;
}
int main()
{
    for(int T = read(); T--; )
    {
        n = read(), ans = 0; for(int i = 1; i <= n; ++i) ea[i].clear(), eb[i].clear(), d[i] = 0;
        for(int i = 1, x, y; i < n; ++i) x = read(), y = read(), ea[x].push_back(y), ea[y].push_back(x), ++d[x], ++d[y];
        for(int i = 1, x, y; i < n; ++i) x = read(), y = read(), eb[x].push_back(y), eb[y].push_back(x);
        dfs(1, 0, ea, pa), dfs(1, 0, eb, pb); for(int i = 1; i <= n; ++i) if(pa[i] != pb[i]) ans = n+1;
        if(ans > n)
            for(int i = 1; i <= n; ++i) if(d[i] == 1)
            {
                dfs(i, 0, eb, pb);
                for(int j = 1; j <= n; ++j) if(i != j)
                {
                    dfs(j, 0, ea, pa), pa[j] = i, pa[i] = 0;
                    int res = solve(); if(~res) ans = min(ans, res+1);
                }
            }            
        printf("%d\n", ans>n?-1:ans);  
    }
    return 0;
}