/*
HDU 6074 - Phone Call [ LCA,并查集 ]  |  2017 Multi-University Training Contest 4
题意：给一棵树，定义集合S(u,v)为u到v路径上所有的点给出 m 个 <S(u1,v1)|S(u2,v2) , w > ，意思为集合里面的点互相距离为 w 求 1 能到的所有点和该生成树的最小权值
分析：将所有线路按代价从小到大排序，对于每条线路(a,b,c,d)分别把S(a,b)和S(c,d)合并到 LCA，最后再把两个 LCA 合并即可再用f(i)表示i向上第一个与i不在同一个连通块的点, 就可用并查集压缩路径
*/
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int N = 100005;
const int Q = 200005;
vector<int> G[N];
namespace LCA{struct Query {int v, q;}; vector <Query> query[N];int ans[Q], f[N], vis[N];int sf(int x) {return x == f[x] ? x : f[x] = sf(f[x]);}void init() {memset(ans, -1, sizeof(ans));for (int i = 0; i < N; i++) {vis[i] = 0;f[i] = i;query[i].clear();}}void adq(int u, int v, int id) {query[u].push_back(Query{v, id});query[v].push_back(Query{u, id});}void LCA(int u) {f[u] = u;vis[u] = 1;for (auto& x : query[u]) {if (vis[x.v] && ans[x.q] == -1)ans[x.q] = sf(x.v);}for (auto& v : G[u]) {if (vis[v]) continue;LCA(v);f[v] = u;}}
}
int pre[N];
void dfs(int u, int fa) {pre[u] = fa;for (auto v : G[u]) {if (v == fa) continue;dfs(v, u);}
}
int f[N], num[N];
LL res[N];
int sf(int x){return x == f[x] ? x : f[x] = sf(f[x]);
}
int same[N];
int Same(int x) {return x == same[x] ? x : same[x] = Same(same[x]);
}
int w;
void join(int a, int b){a = sf(a), b = sf(b);if (a == b) return;f[b] = a;num[a] += num[b];res[a] += res[b] + w;
}
struct Node {int a1, b1, a2, b2, w;int c1, c2;
}e[N];
bool cmp(Node a, Node b) {return a.w < b.w;
}
void solve(int a, int fa) {while (Same(a) != Same(fa)) {a = Same(a);join(pre[a], a);same[a] = pre[a];}
}
int t, n, m;
int main() {scanf("%d", &t);while (t--) {for (int i = 0; i < N; i++) G[i].clear();LCA::init();scanf("%d%d", &n, &m);for (int i = 1; i < n; i++) {int u, v; scanf("%d%d", &u, &v);G[u].push_back(v);G[v].push_back(u);}int Q = 0;for (int i = 1; i <= m; i++) {scanf("%d%d%d%d%d", &e[i].a1, &e[i].b1, &e[i].a2, &e[i].b2, &e[i].w);e[i].c1 = ++Q;LCA::adq(e[i].a1, e[i].b1, Q);e[i].c2 = ++Q;LCA::adq(e[i].a2, e[i].b2, Q);}LCA::LCA(1);for (int i = 1; i <= m; i++) {e[i].c1 = LCA::ans[e[i].c1];e[i].c2 = LCA::ans[e[i].c2];}dfs(1, 1);sort(e+1, e+m+1, cmp);for (int i = 1; i <= n; i++) {f[i] = same[i] = i;num[i] = 1;res[i] = 0;}for (int i = 1; i <= m; i++) {w = e[i].w;solve(e[i].a1, e[i].c1);solve(e[i].b1, e[i].c1);solve(e[i].a2, e[i].c2);solve(e[i].b2, e[i].c2);join(e[i].c1, e[i].c2);}printf("%d %lld\n", num[sf(1)], res[sf(1)]);}
}