点权转边权，把边权设为两个端点的 \(\min\)，然后发现询问 \(x\) 的答案，就是询问 \(x\) 与所有黑点的虚树，边权的 \(\min\) 是多少。假设要判定答案是否 \(\geq k\)，那么就是询问 \(x\) 只经过 \(\geq k\) 是否能到达所有黑点，于是想到建立 Kruskal 重构树，那么 \(x\) 与所有黑点的 LCA 的权值即为答案。时间复杂度为 \(\mathcal{O}(n+q\log n)\)．
严格比 CF1628E 弱，没救了。。。
#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<ctime>
#include<random>
#include<assert.h>
#define pb emplace_back
#define mp make_pair
#define fi first
#define se second
#define dbg(x) cerr<<"In Line "<< __LINE__<<" the "<<#x<<" = "<<x<<'\n'
#define dpi(x,y) cerr<<"In Line "<<__LINE__<<" the "<<#x<<" = "<<x<<" ; "<<"the "<<#y<<" = "<<y<<'\n'
#define DE(fmt,...) fprintf(stderr, "Line %d : " fmt "\n",__LINE__,##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>pii;
typedef pair<ll,int>pli;
typedef pair<ll,ll>pll;
typedef pair<int,ll>pil;
typedef vector<int>vi;
typedef vector<ll>vll;
typedef vector<pii>vpii;
typedef vector<pll>vpll;
template<typename T>T cmax(T &x, T y){return x=x>y?x:y;}
template<typename T>T cmin(T &x, T y){return x=x<y?x:y;}
template<typename T>
T &read(T &r){
r=0;bool w=0;char ch=getchar();
while(ch<'0'||ch>'9')w=ch=='-'?1:0,ch=getchar();
while(ch>='0'&&ch<='9')r=r*10+(ch^48),ch=getchar();
return r=w?-r:r;
}
template<typename T1,typename... T2>
void read(T1 &x,T2& ...y){read(x);read(y...);}
const int N=2000010;
int n,m;
struct Edge{
int x,y,w;
}e[N];
vi eg[N];
int tot,val[N],fa[N];
int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
int dep[N],top[N],pa[N],son[N],siz[N];
void dfs1(int x){
siz[x]=1;
for(auto v:eg[x]){
dep[v]=dep[x]+1;pa[v]=x;
dfs1(v);siz[x]+=siz[v];
if(siz[v]>siz[son[x]])son[x]=v;
}
}
void dfs2(int x,int t){
top[x]=t;
if(son[x])dfs2(son[x],t);
for(auto v:eg[x])if(v!=son[x])dfs2(v,v);
}
int LCA(int x,int y){
while(top[x]!=top[y]){
if(dep[top[x]]<dep[top[y]])swap(x,y);
x=pa[top[x]];
}
return dep[x]<dep[y]?x:y;
}
signed main(){
#ifdef do_while_true
// assert(freopen("data.in","r",stdin));
// assert(freopen("data.out","w",stdout));
#endif
read(n,m);
for(int i=1;i<n;i++)read(e[i].x,e[i].y),e[i].w=min(e[i].x,e[i].y);
sort(e+1,e+n,[&](const Edge &x,const Edge &y){return x.w>y.w;});
tot=n;
for(int i=1;i<=n;i++)fa[i]=val[i]=i;
for(int i=1;i<n;i++){
int x=e[i].x,y=e[i].y,w=e[i].w;
x=getfa(x);y=getfa(y);
++tot;val[tot]=w;
eg[tot].pb(x);eg[tot].pb(y);
fa[x]=fa[y]=fa[tot]=tot;
}
dfs1(tot);
dfs2(tot,tot);
int las=0,t=0;
while(m--){
int op,x;read(op,x);
x=(las+x)%n+1;
if(op==1){
if(!t)t=x;
else t=LCA(t,x);
}
else{
las=val[LCA(t,x)];
cout<<las<<'\n';
}
}
#ifdef do_while_true
// cerr<<'\n'<<"Time:"<<1.0*clock()/CLOCKS_PER_SEC*1000<<" ms"<<'\n';
#endif
return 0;
}