#include <bits/stdc++.h>
using namespace std;

#define lc(p) tr[p].ls
#define rc(p) tr[p].rs

const int N = 3e5 + 10;
struct node {
    int ls, rs, val, siz, rnd;
} tr[N * 20];

int trlen, T, rt[N];

void pushup(int p) {
    tr[p].siz = tr[lc(p)].siz + tr[rc(p)].siz + 1;
}

int newd(node no) {
    tr[++trlen] = no;
    return trlen;
}

int newd(int v) {
    tr[++trlen] = {0, 0, v, 1, rand()};
    return trlen;
}

// 本题按树的大小 siz 划分，可持久化 split 需要在路径上复制节点
void split(int p, int k, int &x, int &y) {
    if (p == 0) {
        x = y = 0;
        return;
    }
    if (tr[lc(p)].siz < k) {
        x = newd(tr[p]); // 复制当前节点
        split(rc(p), k - tr[lc(p)].siz - 1, rc(x), y);
        pushup(x);
    } else {
        y = newd(tr[p]); // 复制当前节点
        split(lc(p), k, x, lc(y));
        pushup(y);
    }
}

int merge(int x, int y) {
    if (!x || !y) return x + y;
    if (tr[x].rnd < tr[y].rnd) {
        rc(x) = merge(rc(x), y);
        pushup(x);
        return x;
    } else {
        lc(y) = merge(x, lc(y));
        pushup(y);
        return y;
    }
}

int getval(int p, int k) {
    if (tr[lc(p)].siz + 1 == k) return tr[p].val;
    if (k <= tr[lc(p)].siz) return getval(lc(p), k);
    else return getval(rc(p), k - tr[lc(p)].siz - 1);
}

int main() {
    int n;
    scanf("%d", &n);
    rt[0] = 0;
    T = trlen = 0;
    
    int x, y, z, t, k, v, op;
    while (n--) {
        scanf("%d", &op);
        if (op == 1) {
            scanf("%d%d%d", &t, &k, &v);
            rt[++T] = rt[t]; // 继承历史版本
            split(rt[T], k - 1, x, y);
            rt[T] = merge(merge(x, newd(v)), y);
        }
        if (op == 2) {
            scanf("%d%d", &t, &k);
            rt[++T] = rt[t]; // 继承历史版本
            // 修正后的删除逻辑：
            split(rt[T], k, x, y);    // x 包含前 k 个元素，y 包含第 k+1 个及以后的元素
            split(x, k - 1, x, z);    // x 包含前 k-1 个元素，z 为第 k 个元素（将被丢弃）
            rt[T] = merge(x, y);      // 合并前 k-1 个和剩余部分，实现删除
        }
        if (op == 3) {
            scanf("%d%d", &t, &k);
            printf("%d\n", getval(rt[t], k));
        }
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define lc(p) tr[p].ls
#define rc(p) tr[p].rs
const int N=3e5+10;
struct node{int ls,rs,val,siz,rnd;}tr[N*20];int trlen,T,rt[N];
void pushup(int p){tr[p].siz=tr[lc(p)].siz+tr[rc(p)].siz+1;}
int newd(node no){tr[++trlen]=no;               return trlen; }
int newd(int   v){tr[++trlen]={0,0,v,1,rand()}; return trlen; }
void split(int p,int k,int &x,int &y)//本题按树的大小siz划分 
{
    if(p==0){x=y=0;return ;}
    if(tr[lc(p)].siz<k)
    {
        x=newd(tr[p]);
        split(rc(p),k-tr[lc(p)].siz-1,rc(x),y);
        pushup(x);
    }
    else 
    {
        y=newd(tr[p]);
        split(lc(p),k,x,lc(y));
        pushup(y);
    }
}

int merge(int x,int y)
{
	if(!x||!y) return x+y;
	if(tr[x].rnd<tr[y].rnd)
	{
		rc(x)=merge(rc(x),y);
		pushup(x);
		return x;
	}
	else
	{
		lc(y)=merge(x,lc(y));
		pushup(y);
		return y;
	}
}
int getval(int p,int k)
{
	if(tr[lc(p)].siz+1==k) return tr[p].val;
	if(k<=tr[lc(p)].siz)   return getval(lc(p),k);
    else                   return getval(rc(p),k-tr[lc(p)].siz-1);
}
int main()
{
    int n;scanf("%d",&n);
    rt[0]=0;T=trlen=0;
    int x,y,z,t,k,v,op;
    while(n--)
	{
        scanf("%d",&op);
        if(op==1)
		{
            scanf("%d%d%d",&t,&k,&v);rt[++T]=rt[t];
            split(rt[T],k-1,x,y);
            rt[T]=merge(merge(x,newd(v)),y);
        }
        if(op==2)
		{
            scanf("%d%d",&t,&k);rt[++T]=rt[t];
            split(rt[T],k,x,y);
            split(x,k-1,x,z);
            rt[T]=merge(x,y);
        }
        if(op==3)
		{
            scanf("%d%d",&t,&k);
            printf("%d\n",getval(rt[t],k));
        }
    }
    return 0;
}