用一个差分树状数组维护一下每个深度的和，然后每次拿着路径端点和lca加一加减一减就行了
#include<bits/stdc++.h>
#define pa pair<int,int>
#define ll long long
using namespace std;
const int maxn=; inline ll rd(){
ll x=;char c=getchar();int neg=;
while(c<''||c>''){if(c=='-') neg=-;c=getchar();}
while(c>=''&&c<='') x=x*+c-'',c=getchar();
return x*neg;
} int N,M,L=;
int eg[maxn*][],egh[maxn],ect;
int dep[maxn],fa[maxn],v[maxn];
int qu[maxn*][],qh[maxn],op[maxn][];
ll tr[maxn];
bool flag[maxn]; inline int lowbit(int x){return x&(-x);}
inline int getfa(int x){return x==fa[x]?x:fa[x]=getfa(fa[x]);}
inline void adeg(int a,int b){
eg[++ect][]=b;eg[ect][]=egh[a];egh[a]=ect;
}inline void adq(int a,int b,int i){
qu[i][]=b;qu[i][]=qh[a];qh[a]=i;
} inline void add(int x,int y){
for(;x&&x<=L;x+=lowbit(x)) tr[x]+=y;
}
inline ll query(int x){
ll re=;for(;x;x-=lowbit(x)) re+=tr[x];return re;
} void tarjan(int x){
flag[x]=;
for(int i=egh[x];i!=-;i=eg[i][]){
if(flag[eg[i][]]) continue;
dep[eg[i][]]=dep[x]+;L=max(L,dep[x]+);
tarjan(eg[i][]);fa[getfa(eg[i][])]=getfa(x);
}
for(int i=qh[x];i!=-;i=qu[i][]){
if(flag[qu[i][]]) op[i>>][]=getfa(qu[i][]);
}
} int main(){
int i,j,k;
N=rd(),M=rd();
memset(egh,-,sizeof(egh));memset(qh,-,sizeof(qh));
for(i=;i<=N;i++) v[i]=rd();
for(i=;i<N;i++){
int a=rd(),b=rd();
adeg(a,b);adeg(b,a);
}for(i=;i<=M;i++){
int a=rd(),b=rd();
if(a==){
int c=rd();adq(b,c,i<<);adq(c,b,i<<|);
op[i][]=b,op[i][]=c;
}else op[i][]=b;
}for(i=;i<=N;i++) fa[i]=i;
dep[]=;tarjan();//printf("ll");
for(i=;i<=N;i++){
add(dep[i],v[i]);add(dep[i]+,-v[i]);
}
for(i=;i<=M;i++){
if(op[i][]){
add(dep[op[i][]],);add(dep[op[i][]]+,-);
if(op[i][]!=op[i][]) add(dep[op[i][]]+,),add(dep[op[i][]]+,-);
if(op[i][]!=op[i][]) add(dep[op[i][]]+,),add(dep[op[i][]]+,-);
}else{
if(op[i][]>L) printf("0\n");
else printf("%lld\n",query(op[i][]));
}
} return ;
}