#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<climits>
#include<cctype>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn=1000005;
int n,m;
struct PerSegmentTree
{
    int a[maxn];
    int val[maxn * 25], cnt; //sum记录原数组区间[1,k]的改编号对应区间值的数量
    int ls[maxn * 25], rs[maxn * 25]; //ls左二子，rs右儿子
    int root[maxn];                   //第i棵线段树的根编号
    PerSegmentTree()
    {
        cnt=0;
    }
    int build(int l, int r)
    {
        int now = ++cnt;
        if (l == r)
        {
            val[now]=a[l];
            return now;
        }
        int mid = (l + r) >> 1;
        ls[now] = build(l, mid);
        rs[now] = build(mid + 1, r);
        return now;
    }
    //将位置y修改为k
    int modify(int x,int l, int r, int y, int k)
    {
        int now = ++cnt;
        if (l == r)
        {
            val[now] = k;
            return now;
        }
        int mid = (l + r) >> 1;
        if (y <= mid)
        {
            ls[now] = modify(ls[x],1,mid,y,k);
            rs[now] = rs[x];
        }
        else
        {
            ls[now] = ls[x];
            rs[now] = modify(rs[x],mid+1,r,y,k);
        }
        return now;
    }
    int query(int x,int l, int r, int y)//询问x版本y位置的值
    {
        if (l == r)
            return val[x];
        int mid = (l + r) >> 1;
        if (y <= mid)
            return query(ls[x],l, mid, y);
        return query(rs[x],mid + 1, r, y);
    }
} pst;

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&pst.a[i]);
    }
    pst.root[0]=pst.build(1,n);
    for(int i=1;i<=m;i++)
    {
        int ver,op,loc,v;
        scanf("%d%d%d",&ver,&op,&loc);
        if(op==1)
        {
            scanf("%d",&v);
            pst.root[i]=pst.modify(pst.root[ver],1,n,loc,v);
        }
        else
        {
            printf("%d\n",pst.query(pst.root[ver],1,n,loc));
            pst.root[i]=pst.root[ver];
        }
    }
}