#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
int N,Q, root, mn[MX], par[MX][17], key[MX], rkey[MX], p[MX], oc[MX];
vi child[MX];
set<int> lef;
void dfs(int a) {
mn[a] = a;
for (int i: child[a]) {
dfs(i);
mn[a] = min(mn[a],mn[i]);
}
sort(all(child[a]),[](int x, int y) { return mn[x] < mn[y]; });
}
int nex = 0;
int dfs2(int a) {
vi v;
for (int i: child[a]) v.pb(dfs2(i));
key[++nex] = a;
rkey[a] = nex;
for (int i: v) par[i][0] = nex;
// cout << nex << " " << a << "\n";
return nex;
}
void genTree() {
dfs(root);
dfs2(root);
FOR(j,1,17) FOR(i,1,N+1) par[i][j] = par[par[i][j-1]][j-1];
}
int ad() {
int x = *lef.begin();
oc[x] = 1;
lef.erase(x);
return key[x];
}
int rem(int x) {
int cur = rkey[x], ans = 0;
F0Rd(i,17) if (oc[par[cur][i]]) {
cur = par[cur][i];
ans ^= 1<<i;
}
// cout << "AH " << x << " " << cur << "\n";
oc[cur] = 0; lef.insert(cur);
return ans;
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> Q;
FOR(i,1,N+1) {
cin >> p[i];
if (p[i] == 0) root = i;
else child[p[i]].pb(i);
}
genTree();
FOR(i,1,N+1) lef.insert(i);
F0R(i,Q) {
int t,x; cin >> t >> x;
if (t == 1) {
int ans = 0;
F0R(i,x) ans = ad();
cout << ans << "\n";
} else {
cout << rem(x) << "\n";
}
}
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
