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 = 60;

int n, rt;

namespace WEN {

vector<int> G[N]; int ind[N];

inline void adde(int from, int to) {
//    cout << "adde: " << from << ' ' << to << endl;
    ind[to]++, G[from].push_back(to);
}

inline bool check() {
    queue<int> q; int tot = 0;
    for (int i = 1; i <= n; i++) if (!ind[i]) q.push(i);
    while (!q.empty()) {
        int u = q.front(); q.pop(); tot++;
        for (auto v : G[u])
            if (!--ind[v]) q.push(v);
    }
    for (int i = 1; i <= n; i++) G[i].clear(), ind[i] = 0;
    return tot == n;
}

}

vector<int> G1[N], G2[N]; int G[N][N], mark[N], tot, fuck[N][N];

void dfs(int u, int f, int ok) {
    if (ok) mark[u] = 1, tot++;
    else if (!mark[f]) WEN::adde(u, f);
    for (auto v : G1[u]) if (v != f) 
        dfs(v, u, ok & G[u][v]);
}

void dfs1(int u, int f) {
    if (!mark[u] && !mark[f]) WEN::adde(f, u);
    mark[u] = 0;
    for (auto v : G2[u]) if (v != f) dfs1(v, u);
}

inline void erase(vector<int> &t, int a) {
    for (auto &v : t) if (v == a) {
        swap(v, t.back()), t.pop_back(); return;
    }
    assert(0);
}

int main() {
    int T; scanf("%d", &T);
    while (T--) {
        int res = 1e9; scanf("%d", &n);
        memset(G, 0, sizeof(G)), memset(fuck, 0, sizeof(fuck));
        for (int i = 1; i <= n; i++) G1[i].clear(), G2[i].clear();
        for (int i = 1, a, b; i < n; i++)
            scanf("%d%d", &a, &b), G1[a].push_back(b), G1[b].push_back(a), fuck[a][b] = fuck[b][a] = 1;
        for (int i = 1, a, b; i < n; i++)
            scanf("%d%d", &a, &b), G2[a].push_back(b), G2[b].push_back(a), G[a][b] = G[b][a] = 1;
        for (int i = 1; i <= n; i++) if (G1[i].size() == 1)
            for (int j = 1; j <= n; j++) if (j != i) {
                int v = G1[i].back(); G1[i].pop_back(), erase(G1[v], i);
                G1[i].push_back(j), G1[j].push_back(i), tot = 0;
                rt = i, dfs(i, 0, 1), dfs1(i, 0);
                G1[i].pop_back(), G1[i].push_back(v), G1[j].pop_back(), G1[v].push_back(i);
                if (WEN::check()) res = min(res, n - tot + !fuck[i][j]);
//                cout << "check: " << i << ' ' << j << ' ' << res << endl;
            }
        if (res <= n) printf("%d\n", res);
        else puts("-1");
    }
}