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 = 55, inf = 1e9;
int nTest, n, ans;
bool vis[N], canUse[N];
vector< set<int> > gr[3];
set<int> inQ;

void dfs (int u) {
    vis[u] = 1;
    for (auto v : gr[2][u]) {
        if (vis[v]) continue ;
        if (gr[0][u].count(v) ) dfs(v);
        else inQ.insert(v);
    }
}

int solve (int root) {
    memset(vis, 0, sizeof vis);
    memset(canUse, 0, sizeof canUse);
    inQ.clear();
    int ret = 0;
    gr[0] = gr[1];
    for (int i = 1; i <= n; ++i) if (gr[0][i].size() == 1) canUse[i] = 1;
    dfs(root);
    while (1) {
        bool have = 0;
        for (auto i : inQ) if (canUse[i]) {
            gr[0][ *gr[0][i].begin() ].erase(i);
            gr[0][i].clear();
            have = 1, dfs(i), inQ.erase(i), ret ++;
            for (int j = 1; j <= n; ++j) if (gr[0][j].size() == 1) canUse[j] = 1;
            break;
        }
        if (!inQ.size() ) break;
        if (!have) return inf;
    }
    return ret;
}

void EntityIT() {
    cin >> n;
    ans = inf;
    for (int i = 1; i < 3; ++i) gr[i].assign(n + 5, set<int>() );
    for (int type = 1; type < 3; ++type) {
        for (int _ = 1; _ < n; ++_) {
            int u, v; cin >> u >> v;
            gr[type][u].insert(v); gr[type][v].insert(u);
        }
    }

    for (int root = 1; root <= n; ++root) ans = min(ans, solve(root) );
    for (int root = 1; root <= n; ++root) {
        if (gr[1][root].size() > 1) continue ;
        for (int i = 1; i <= n; ++i) {
            if (gr[1][root].count(i) || i == root) continue ;
            int tmp = (*gr[1][root].begin() );
            gr[1][i].insert(root);
            gr[1][root].clear(); gr[1][root].insert(i);
            gr[1][tmp].erase(root);
            ans = min(ans, solve(root) + 1);
            gr[1][i].erase(root);
            gr[1][root].clear(); gr[1][root].insert(tmp);
            gr[1][tmp].insert(root);
        }
    }

    if (ans >= inf) cout << "-1\n";
    else cout << ans << '\n';
}

int main () {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    //freopen("test.inp", "r", stdin);
    cin >> nTest;
    while (nTest --) EntityIT();
    return 0;
}