T1
简单题，用 dfs 序拍扁树，线性基可以维护 xor kth，套棵线段树实现点修区间查询即可。
#include<bits/stdc++.h>
#define N 50005
using namespace std;
int n,q,g,tot;
vector<int>s[N];
int a[N],dfn[N],siz[N];
class SGT{#define l(i) ((i)<<1)#define r(i) ((i)<<1|1)private:class node{private:int a[20];int rebuild(){int tmp=0;for(int i=0;i<20;i++)for(int j=i-1;~j;j--)if(a[i]&1<<j) a[i]^=a[j];for(int i=0;i<20;i++)if(a[i]) a[tmp++]=a[i];return tmp;}public:node(){memset(a,0,sizeof a);}node(int x):node(){insert(x);}node operator+(node x){return x+=*this,x;}void operator+=(const node &x){for(int i=0;i<20;i++)if(x.a[i]) insert(x.a[i]);}void insert(int x){for(int i=19;~i;i--) if(x&1<<i)if(!a[i]) return a[i]=x,void();else x^=a[i];}int kth(int x){int t=rebuild();if(x>1<<t) return -1;int res=0;x--;for(int i=0;i<t;i++)if(x&1<<i) res^=a[i];return res;}}tr[N<<2];public:void upd(int p,int v,int x=1,int l=1,int r=n){ if(l==r) return tr[x]=v,void();int mid=l+r>>1;if(p<=mid) upd(p,v,l(x),l,mid);else upd(p,v,r(x),mid+1,r);tr[x]=tr[l(x)]+tr[r(x)];}node query(int ql,int qr,int x=1,int l=1,int r=n){if(ql<=l&&qr>=r) return tr[x];int mid=l+r>>1;node res;if(ql<=mid) res+=query(ql,qr,l(x),l,mid);if(qr>mid) res+=query(ql,qr,r(x),mid+1,r);return res;}#undef l#undef r
}tr;
void dfs(int x,int fa){tr.upd(dfn[x]=++tot,a[x]),siz[x]=1;for(auto p:s[x]) if(p^fa)dfs(p,x),siz[x]+=siz[p];
}
int main(){scanf("%d%d",&n,&g);for(int i=1,x,y;i<n;i++)scanf("%d%d",&x,&y),s[x].emplace_back(y),s[y].emplace_back(x);for(int i=1;i<=n;i++)scanf("%d",a+i);scanf("%d",&q),dfs(1,0);while(q--){int op,x,k;scanf("%d%d%d",&op,&x,&k);if(op==1) tr.upd(dfn[x],k);else printf("%d\n",tr.query(dfn[x],dfn[x]+siz[x]-1).kth(k));}return 0;
}