/*[продолжатьипродолжать]*/
/* ---------------------- *
* Code by watbor *
* *
--------------------------*/
#define tct template<class MITSURII,class CHANN>
#include <bits/stdc++.h>
using namespace std ;
#define ll long long
#define fi first
#define se second
#define pb push_back
#define ii pair<int,int>
#define vii vector<ii>
#define vi vector<int>
#define endl '\n'
#define sz(a) (int)a.size()
#define all(a) a.begin(),a.end()
#define FOR(i,a,b) for(int i(a) ; i <= b ; i++)
#define FORD(i,a,b) for(int i(a) ; i >= b ; i--)
#define FORN(i,a,b) for(int i(a) ; i < b ; i++)
#define uni(a) sort(all(a)) , a.resize(unique(a.begin(),a.end())-a.begin())
const int N = 2e5 + 5 ;
const int LO = __lg(N) + 1 ;
const int base = 311 ;
const int M = 1e3 + 5 ;
const int oo = 2e9 ;
const int mod = 1e9 + 7 ;
const long long inf = 1e18 ;
const int dx[] = {-1,1,0,0} ;
const int dy[] = {0,0,-1,1} ;
tct bool minimize(MITSURII &x,const CHANN &y) {if(x > y) {x = y ; return true ; }else return false ;}
tct bool maximize(MITSURII &x,const CHANN &y) {if(x < y) {x = y ; return true ; }else return false ;}
/* Solution : Set + LCA
để ý rằng đáp án luôn tồn tại ở dạng (i,i+1) vì a[i] với a[i+1]
bởi ít nhất phải thuộc 2 cây con khác nhau gốc i
-> chỉ cần quan tâm tới 2 thằng liên tiếp
*/
int n , m , q ;
int a[N] ;
vi g[N] ;
void init() {
cin >> n >> m >> q ;
FOR(i,1,n-1) {
int u , v ; cin >> u >> v ;
g[u].pb(v) ; g[v].pb(u) ;
}
FOR(i,1,m) cin >> a[i] ;
}
struct {
int P[N][LO] , h[N] ;
void dfs(int u,int par) {
for(auto v : g[u]) if(v!=par) {
h[v] = h[u] + 1 ;
P[v][0] = u ;
dfs(v,u) ;
}
}
void init() {
dfs(1,1) ;
for(int j(1) ; (1<<j) <= n ; j++) {
FOR(i,1,n) {
int x = P[i][j-1] ;
P[i][j] = P[x][j-1] ;
}
}
}
int find(int u,int v) {
if(u==v) return u ;
if(h[u]<h[v])swap(u,v) ;
int z = __lg(h[u]) ;
FORD(s,z,0)if((h[u]-h[v])>=(1<<s)) {
u = P[u][s] ;
}
if(u==v)return u ;
FORD(s,z,0)if(P[u][s]!=P[v][s]) {
u = P[u][s] ;
v = P[v][s] ;
}
return P[u][0] ;
}
} LCA ;
set<int> pos[N] , lca[N] ;
void solve() {
LCA.init() ;
FOR(i,1,m) pos[a[i]].insert(i) ;
FOR(i,1,m-1) lca[LCA.find(a[i],a[i+1])].insert(i) ;
while(q--) {
int t ; cin >> t ;
if(t==1) {
int i , u ;
cin >> i >> u ;
pos[a[i]].erase(i) ;
pos[u].insert(i) ;
if(i < m) {
int old_LCA = LCA.find(a[i],a[i+1]) ;
lca[old_LCA].erase(i) ;
}
if(i >= 2) {
int old_LCA = LCA.find(a[i],a[i-1]) ;
lca[old_LCA].erase(i-1) ;
}
a[i] = u ;
if(i < m) {
int new_LCA = LCA.find(a[i],a[i+1]) ;
lca[new_LCA].insert(i) ;
}
if(i >= 2) {
int new_LCA = LCA.find(a[i],a[i-1]) ;
lca[new_LCA].insert(i-1) ;
}
} else {
int l , r , u ;
cin >> l >> r >> u ;
int cur_val = 0 ;
if(!pos[u].empty()) {
auto it = pos[u].lower_bound(l) ;
cur_val = *it ;
if(! ( cur_val >= l && cur_val <= r) ) cur_val = 0 ;
}
if(cur_val) cout << cur_val << ' ' << cur_val << '\n' ;
else {
auto it = lca[u].lower_bound(l) ;
cur_val = *it ;
if(cur_val >= l && cur_val < r) cout << cur_val << ' ' << cur_val + 1 << '\n' ;
else cout << -1 << ' ' << -1 << '\n' ;
}
}
}
}
signed main() {
#define TASK "ccc"
ios_base::sync_with_stdio(0);
cin.tie(nullptr);cout.tie(nullptr);
if( fopen(TASK".inp","r") ) {
freopen(TASK".inp","r",stdin) ; freopen(TASK".out","w",stdout);
}
init() ;
solve() ;
cerr << endl << "watborhihi : " << clock() << "ms" << endl;
return 0 ;
}
