// Skyqwq
#include <bits/stdc++.h>

#define pb push_back
#define fi first
#define se second
#define mp make_pair

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }

template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

const int N = 2e5 + 5;

int n, m, dfn[N], fa[N], dfncnt, L[N], R[N], sz[N];

vector<int> g[N], w[N];

struct BIT{
	int c[N];
	void inline add(int x, int k) {
		for (; x <= n; x += x & -x) c[x] += k;
	}

	int inline ask(int x) {
		int ret = 0;
		for (; x; x -= x & -x) ret += c[x];
		return ret;
	}
	void inline clr() {
		for (int i = 1; i <= n; i++) c[i] = 0;
	}
} t1, t2;

struct E{
	int s, t;
} e[N];


bool vis[N];

int fr[N];

void inline clr() {
	dfncnt = 0;
	t1.clr(), t2.clr();
	for (int i = 1; i <= n; i++) vis[i] = fr[i] = 0, g[i].clear(), w[i].clear(), fa[i] = sz[i] = dfn[i] = 0;
}

void dfs1(int u) {
	dfn[u] = ++dfncnt;
	sz[u] = 1;
	L[u] = dfn[u];
	for (int v: g[u]) {
		if (v == fa[u]) continue;
		fa[v] = u;
		dfs1(v);
		sz[u] += sz[v];
	}
	R[u] = dfncnt;
}

bool inline cmp(E x, E y) {
	if (dfn[x.s] != dfn[y.s]) return dfn[x.s] < dfn[y.s];
	return dfn[x.t] > dfn[y.t];
}

bool dfs2(int u) {
	bool o = 0;
	for (int v: g[u]) {
		if (v == fa[u]) continue;
		o |= dfs2(v);
	}
	if (!o && vis[u]) t1.add(dfn[u], 1);
	return o | vis[u];
}

void solve() {
	int p = 1, re = 0, ans = 0;
	while (p) {
		int sum = 0, mx = 0, big = 0;
		for (int v: g[p]) {
			if (v != fa[p]) {
				int w = t1.ask(R[v]) - t1.ask(L[v] - 1);
				if (chkMax(mx, w)) big = v;
				sum += w;
			}
		}
		int rt = sum - mx;
		int rs = rt + re;
		if (mx <= rs) {
			ans += sum - (sum + re) / 2;
			break;
		}
		ans += rt;
		re += rt;
		for (int v: w[p]) {
			if (dfn[v] < L[big] || dfn[v] > R[big]) continue;
			int z = t2.ask(dfn[v]);
			if (z) {
				t1.add(dfn[z], 1);
				t2.add(L[z], -z), t2.add(R[z] + 1, z);
				fr[z]--;
			} else if (re) {
				re--;
			} else {
				ans++;
			}
			t1.add(dfn[v], -1);
			t2.add(L[v], v), t2.add(R[v] + 1, -v);
			fr[v]++;
		}
		if (fr[big]) t2.add(L[big], -big), t2.add(R[big] + 1, big);
		p = big;
		re += fr[big], fr[big] = 0;
	}
	printf("%d\n", ans);
}

void inline work() {
	read(n), read(m);
	for (int i = 1, u, v; i < n; i++) 
		read(u), read(v), g[u].pb(v), g[v].pb(u);
	for (int i = 1; i <= m; i++) read(e[i].s), read(e[i].t);
	dfs1(1);
	// 清除包含关系
	sort(e + 1, e + 1 + m, cmp);
	for (int i = 1; i <= m; i++) {
		int cnt = t1.ask(R[e[i].t]) - t1.ask(L[e[i].t] - 1);
		if (!cnt) {
			w[e[i].s].pb(e[i].t);
			vis[e[i].t] = 1;
		}
		t1.add(dfn[e[i].t], 1);
	}
	t1.clr();
	// 维护 t1，维护在最底下、未解决的 t (s 在当前子树里)
	dfs2(1);
	// 维护 t2，维护那些互不成祖先关系的特殊点，覆盖他们的子树，帮助快速查
	solve();
	clr();
}

int main() {
	// freopen("memory6.in", "r", stdin);
	// freopen("m.out", "w", stdout);
    int Case; read(Case);
    while (Case--) work();
    return 0;
}