#include <bits/stdc++.h>
#define FOR(i, a, b) for(int i = (a); i <= (b); i++)
#define FOD(i, a, b) for(int i = (a); i >= (b); i--)
#define file(name) if(fopen(name".inp", "r")) { freopen(name".inp", "r", stdin); freopen(name".out", "w", stdout); }
#define fi first
#define se second
#define el cout << '\n'
#define maxn int(1e5 + 5)
using namespace std;
typedef pair<int, int> pii;
const int LOG = 17;
int n, m, q;
int a[maxn], b[maxn], c[maxn];
int d[maxn], par[maxn][20];
vector<int> adj[maxn];
void dfs(int u) {
for(int v : adj[u]) if(v != par[u][0]) {
d[v] = d[u] + 1;
par[v][0] = u;
dfs(v);
}
}
int lca(int u, int v) {
if(d[u] < d[v]) swap(u, v);
int k = d[u] - d[v];
FOD(i, LOG, 0) if(k >> i & 1) u = par[u][i];
if(u == v) return u;
FOD(i, LOG, 0) if(par[u][i] != par[v][i]) u = par[u][i], v = par[v][i];
return par[u][0];
}
namespace subtask2 {
bool check() {
return n <= 2000 && m <= 2000 && q <= 2000;
}
int diff[2005];
void dfscalc(int u) {
for(int v : adj[u]) if(v != par[u][0]) {
dfscalc(v);
diff[u] += diff[v];
}
}
void update(int u, int v) {
int l = lca(u, v);
diff[u]++;
diff[v]++;
diff[l]--;
diff[par[l][0]]--;
}
void solve() {
while(q--) {
FOR(i, 1, n) diff[i] = 0;
int l, r;
cin >> l >> r;
if(l == r) {
cout << 1, el;
continue;
}
FOR(i, l + 1, r) update(c[i - 1], c[i]);
dfscalc(1);
int res = 0;
FOR(i, 1, n) if(diff[i]) res++;
cout << res, el;
}
}
}
namespace subtask3 {
bool check() {
FOR(i, 1, n - 1) if(a[i] != i || b[i] != i + 1) return 0;
return 1;
}
pii st[maxn][20];
pii combine(pii &A, pii &B) {
return { min(A.fi, B.fi), max(A.se, B.se) };
}
pii get(int l, int r) {
int k = __lg(r - l + 1);
return combine(st[l][k], st[r - (1 << k) + 1][k]);
}
void solve() {
FOR(i, 1, m) st[i][0] = {c[i], c[i]};
FOR(j, 1, LOG)
FOR(i, 1, n - (1 << j) + 1)
st[i][j] = combine(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
while(q--) {
int l, r;
cin >> l >> r;
pii res = get(l, r);
cout << res.se - res.fi + 1, el;
}
}
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin >> n >> m >> q;
FOR(i, 1, n - 1) {
cin >> a[i] >> b[i];
adj[a[i]].push_back(b[i]);
adj[b[i]].push_back(a[i]);
}
FOR(i, 1, m) cin >> c[i];
dfs(1);
FOR(i, 1, LOG)
FOR(u, 1, n)
par[u][i] = par[par[u][i - 1]][i - 1];
if(subtask2::check()) return subtask2::solve(), 0;
if(subtask3::check()) return subtask3::solve(), 0;
return 0;
}
