Problem: Given a tree with n (1 ≤ n ≤ 200,000) nodes and a list of q (1 ≤ q ≤ 100,000) queries, process the queries in order and output a value for each output query. The given tree is connected and each node on the tree has a weight wi (-10,000 ≤ wi ≤ 10,000).

Each query consists of a number ti (ti = 1, 2), which indicates the type of the query , and three numbers ai, bi and ci (1 ≤ ai, bi ≤ n, -10,000 ≤ ci ≤ 10,000). Depending on the query type, process one of the followings:

* (ti = 1: modification query) Change the weights of all nodes on the shortest path between ai and bi (both inclusive) to ci.

* (ti = 2: output query) First, create a list of weights on the shortest path between ai and bi (both inclusive) in order. After that, output the maximum sum of a non-empty continuous subsequence of the weights on the list. ci is ignored for output queries.



Input

The first line contains two integers n and q. On the second line, there are n integers which indicate w1, w2, ... , wn.

Each of the following n - 1 lines consists of two integers si and ei (1 ≤ si, ei ≤ n), which means that there is an edge between si and ei.

Finally the following q lines give the list of queries, each of which contains four integers in the format described above. Queries must be processed one by one from top to bottom.

Output

For each output query, output the maximum sum in one line.

Examples

Input

3 4
1 2 3
1 2
2 3
2 1 3 0
1 2 2 -4
2 1 3 0
2 2 2 0


Output

6
3
-4


Input

7 5
-8 5 5 5 5 5 5
1 2
2 3
1 4
4 5
1 6
6 7
2 3 7 0
2 5 2 0
2 4 3 0
1 1 1 -1
2 3 7 0


Output

12
10
10
19


Input

21 30
10 0 -10 -8 5 -5 -4 -3 1 -2 8 -1 -7 2 7 6 -9 -6 3 4 9
10 3
3 2
3 12
12 4
4 13
4 9
10 21
21 1
1 11
11 14
1 15
10 6
6 17
6 16
6 5
5 18
5 19
10 7
10 8
8 20
1 1 21 -10
1 3 19 10
2 1 13 0
1 4 18 8
1 5 17 -5
2 16 7 0
1 6 16 5
1 7 15 4
2 4 20 0
1 8 14 3
1 9 13 -1
2 9 18 0
1 10 12 2
1 11 11 -8
2 21 15 0
1 12 10 1
1 13 9 7
2 6 14 0
1 14 8 -2
1 15 7 -7
2 10 2 0
1 16 6 -6
1 17 5 9
2 12 17 0
1 18 4 6
1 19 3 -3
2 11 8 0
1 20 2 -4
1 21 1 -9
2 5 19 0


Output

20
9
29
27
10
12
1
18
-2
-3

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using Int = long long;
#include<bits/stdc++.h>
using namespace std;
using Int = long long;
 
struct HLDecomposition {
  int n,pos;
  vector<vector<int> > G;
  vector<int> vid, head, sub, hvy, par, dep, inv, type;
   
  HLDecomposition(){}
  HLDecomposition(int sz):
    n(sz),pos(0),G(n),
    vid(n,-1),head(n),sub(n,1),hvy(n,-1),
    par(n),dep(n),inv(n),type(n){}
   
  void add_edge(int u, int v) {
    G[u].push_back(v);
    G[v].push_back(u);
  }
 
  void build(vector<int> rs={0}) {
    int c=0;
    for(int r:rs){
      dfs(r);
      bfs(r, c++);
    }
  }
   
  void dfs(int rt) {
    using T = pair<int, int>;
    stack<T> st;
    par[rt]=-1;
    dep[rt]=0;
    st.emplace(rt,0);
    while(!st.empty()){
      int v=st.top().first;
      int &i=st.top().second;
      if(i<(int)G[v].size()){
    int u=G[v][i++];
    if(u==par[v]) continue;
    par[u]=v;
    dep[u]=dep[v]+1;
    st.emplace(u,0);
      }else{
    st.pop();
    int res=0;
    for(int u:G[v]){
      if(u==par[v]) continue;
      sub[v]+=sub[u];
      if(res<sub[u]) res=sub[u],hvy[v]=u;
    }
      }
    }
  }
 
  void bfs(int r,int c) {
    int &k=pos;
    queue<int> q({r});
    while(!q.empty()){
      int h=q.front();q.pop();
      for(int i=h;i!=-1;i=hvy[i]) {
    type[i]=c;
    vid[i]=k++;
    inv[vid[i]]=i;
    head[i]=h;
    for(int j:G[i])
      if(j!=par[i]&&j!=hvy[i]) q.push(j);
      }
    }
  }
   
  // for_each(vertex)
  // [l,r] <- attention!!
  void for_each(int u, int v, const function<void(int, int)>& f) {
    while(1){
      if(vid[u]>vid[v]) swap(u,v);
      f(max(vid[head[v]],vid[u]),vid[v]);
      if(head[u]!=head[v]) v=par[head[v]];
      else break;
    }
  }
   
  // for_each(edge)
  // [l,r] <- attention!!
  void for_each_edge(int u, int v, const function<void(int, int)>& f) {
    while(1){
      if(vid[u]>vid[v]) swap(u,v);
      if(head[u]!=head[v]){
    f(vid[head[v]],vid[v]);
        v=par[head[v]];
      } else{
    if(u!=v) f(vid[u]+1,vid[v]);
    break;
      }
    }
  }
 
  int lca(int u,int v){
    while(1){
      if(vid[u]>vid[v]) swap(u,v);
      if(head[u]==head[v]) return u;
      v=par[head[v]];
    }
  }
 
  int distance(int u,int v){
    return dep[u]+dep[v]-2*dep[lca(u,v)];
  }
};
 
 
template <typename T,typename E>
struct SegmentTree{
  using F = function<T(T,T)>;
  using G = function<T(T,E)>;
  using H = function<E(E,E)>;
  using P = function<E(E,size_t)>;
  int n;
  F f;
  G g;
  H h;
  T ti;
  E ei;
  P p;
  vector<T> dat;
  vector<E> laz;
  SegmentTree(int n_,F f,G g,H h,T ti,E ei,
	      P p=[](E a,size_t b){b++;return a;}):
    f(f),g(g),h(h),ti(ti),ei(ei),p(p){
    init(n_);
  }
  void init(int n_){
    n=1;
    while(n<n_) n*=2;
    dat.assign(2*n-1,ti);
    laz.assign(2*n-1,ei);
  }
  void build(int n_, vector<T> v){
    for(int i=0;i<n_;i++) dat[i+n-1]=v[i];
    for(int i=n-2;i>=0;i--)
      dat[i]=f(dat[i*2+1],dat[i*2+2]);
  }
  inline void eval(int len,int k){
    if(laz[k]==ei) return;
    if(k*2+1<n*2-1){
      laz[k*2+1]=h(laz[k*2+1],laz[k]);
      laz[k*2+2]=h(laz[k*2+2],laz[k]);
    }
    dat[k]=g(dat[k],p(laz[k],len));
    laz[k]=ei;
  }
  T update(int a,int b,E x,int k,int l,int r){
    eval(r-l,k);
    if(r<=a||b<=l) return dat[k];
    if(a<=l&&r<=b){
      laz[k]=h(laz[k],x);
      return g(dat[k],p(laz[k],r-l));
    }
    return dat[k]=f(update(a,b,x,k*2+1,l,(l+r)/2),
		    update(a,b,x,k*2+2,(l+r)/2,r));
  }
  T update(int a,int b,E x){
    return update(a,b,x,0,0,n);
  }
  T query(int a,int b,int k,int l,int r){
    eval(r-l,k);
    if(r<=a||b<=l) return ti;
    if(a<=l&&r<=b) return dat[k];
    T vl=query(a,b,k*2+1,l,(l+r)/2);
    T vr=query(a,b,k*2+2,(l+r)/2,r);
    return f(vl,vr);
  }
  T query(int a,int b){
    return query(a,b,0,0,n);
  }
  void update(int k,T x){
    query(k,k+1);//evaluate
    k+=n-1;
    dat[k]=x;
    while(k){
      k=(k-1)/2;
      dat[k]=f(dat[k*2+1],dat[k*2+2]);
    }
  }
};

//INSERT ABOVE HERE
signed main(){
  int n,q;
  scanf("%d %d",&n,&q);
  HLDecomposition hld(n);
  vector<int> w(n);
  for(int i=0;i<n;i++) cin>>w[i];
  using P = pair<int,int>;
  for(int i=0;i<n-1;i++){
    int a,b;
    cin>>a>>b;
    a--;b--;
    hld.add_edge(a,b);
  }
  hld.build();
  
  using T = tuple<int,int,int,int,int,int,int>;
  
  T d1(-1,-1,-1,-1,-1,-1,-1);
  P d0(-1,-114514);

  auto &par=hld.par;
  auto &vid=hld.vid;;
  auto con=[&](int a,int b){
    return par[a]==b||par[b]==a;
  };
  
  auto f=[&](T a,T b){
    if(a>b) swap(a,b);
    
    if(get<0>(a)<0) return b;
    if(con(get<0>(a),get<1>(b))) swap(a,b);
   
    int al,ar,as,ava,avi,avl,avr;
    tie(al,ar,as,ava,avi,avl,avr)=a;
    int bl,br,bs,bva,bvi,bvl,bvr;
    tie(bl,br,bs,bva,bvi,bvl,bvr)=b;
    
    if(!con(ar,bl)){
      if(con(ar,br)){
	swap(bl,br);
	swap(bvl,bvr);
      }else if(con(al,bl)){
	swap(al,ar);
	swap(avl,avr);
      }else{
	return d1;
      }
    }
    int cl=al,cr=br,cs=as+bs;
    int cva=ava+bva,cvi=max(avi,bvi),cvl=avl,cvr=bvr;
    cvi=max(cvi,avr+bvl);
    cvl=max(cvl,ava+bvl);
    cvr=max(cvr,avr+bva);
    
    return T(cl,cr,cs,cva,cvi,cvl,cvr);
  };
  
  auto g=[&](T a,P p){
    if(p==d0) return a;
    int al,ar,as,ava,avi,avl,avr;
    tie(al,ar,as,ava,avi,avl,avr)=a;
    int v=p.first,b=p.second;
    if(~v) al=ar=v,as=1;
    if(b>=0) return T(al,ar,as,b*as,b*as,b*as,b*as);
    return T(al,ar,as,b*as,b,b,b);
  };
  
  auto h=[&](P a,P b){a.first++;return b;};

  vector<T> vt(n);
  for(int i=0;i<n;i++)
    vt[vid[i]]=g(d1,P(i,w[i]));

  SegmentTree<T,P> seg(n,f,g,h,d1,d0);
  seg.build(n,vt);
  
  while(q--){
    int t,a,b,c;
    scanf("%d %d %d %d",&t,&a,&b,&c);
    a--;b--;
    if(t==1){
      hld.for_each(a,b,[&](int l,int r){
	  seg.update(l,r+1,P(-1,c));
	});
    }
    if(t==2){
      vector<T> v;
      hld.for_each(a,b,[&](int l,int r){
	  T x=seg.query(l,r+1);
	  for(int i=0;i<(int)v.size();i++){
	    T y=f(v[i],x);
	    if(~get<0>(y)){
	      v[i]=y;
	      return;
	    }
	  }
	  v.push_back(x);
	});
      while(v.size()>1){
	T x=v.back();v.pop_back();
	for(int i=0;i<(int)v.size();i++){
	  T y=f(v[i],x);
	  if(~get<0>(y)){
	    v[i]=y;
	    break;
	  }
	}
      }
      int vl,vr,vs,vva,vvi,vvl,vvr;
      tie(vl,vr,vs,vva,vvi,vvl,vvr)=v.front();
      printf("%d\n",max({vva,vvi,vvl,vvr}));
    }
  }
  
  return 0;
}