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 ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

struct HL_decomposition{
    int n;
    vector<vector<int>> G;
    vector<int> vid, inv, depth, par, heavy, head, sub;
    /*
    vid : HL分解後のグラフでの頂点ID
    inv : HL分解前のグラフでのvid[i]の頂点ID
    depth : rootからの距離
    par : 根付き木上での親
    heavy : heavy-path上における頂点iの次の頂点ID
    head : 頂点iが属するheavy-pathの先頭の頂点ID
    sub : 部分木のサイズ
    */

    HL_decomposition(){}
    HL_decomposition(int sz) : n(sz), G(n), vid(n), inv(n), depth(n), par(n), heavy(n,-1), head(n), sub(n) {}

    void add_edge(int u, int v){
        G[u].push_back(v);
        G[v].push_back(u);
    }

    void build(int root = 0){
        dfs(root, -1);
        dfs2(root);
    }

    void dfs(int cur, int pre){
        par[cur] = pre;
        sub[cur] = 1;
        int max_sub = 0;
        for(int nx:G[cur])if(nx != pre){
            depth[nx] = depth[cur] + 1;
            dfs(nx, cur);

            sub[cur] += sub[nx];
            if(max_sub < sub[nx]){
                max_sub = sub[nx];
                heavy[cur] = nx;
            }
        }
    }

    void dfs2(int root){
        int k = 0;
        stack<int> que({root});
        while(!que.empty()){
            int cur = que.top();
            que.pop();
            // curを先頭とするheavy-pathを処理
            for(int i=cur; i!=-1; i=heavy[i]){
                vid[i] = k++;
                inv[vid[i]] = i;
                head[i] = cur;
                // heavy-pathの先頭になるなら、queにpushする
                for(int nx:G[i])if(nx != par[i] && nx != heavy[i]){
                    que.push(nx);
                }
            }
        }
    }

    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]];
        }
    }
};

HL_decomposition hl;

struct Seg{
    // 左端 / 右端になる時のbest
    ll l,r;
    // 全体の和
    ll w;

    // 部分和の最大値
    ll mx;
    // mxが0になる時、空にならないためマシな値を持つ
    ll nb;

    // 区間内の要素数
    int sz;

    Seg(){
        sz = 0;
    }
    Seg(int val){
        sz = 1;
        w = val;

        l = r = mx = 0;
        if(val>=0) l = r = mx = val;

        nb = val;
    }

    void show(){
        printf("l,r,w %lld %lld %lld  mx %lld  nb %lld, sz(%d)\n",l,r,w,mx,nb,sz);
    }
};

Seg MERGE(Seg lh, Seg rh){
    if(lh.sz == 0) return rh;
    if(rh.sz == 0) return lh;

    Seg ret;
    ret.sz = lh.sz + rh.sz;

    ret.l = max(rh.l, lh.l + rh.w);
    if(ret.l < 0) ret.l = 0;

    ret.r = max(lh.r, lh.w + rh.r);
    if(ret.r < 0) ret.r = 0;

    ret.w = lh.w + rh.w;

    ret.mx = max({lh.mx, rh.mx, lh.l+rh.r});
    ret.nb = max(lh.nb, rh.nb);
    return ret;
}

Seg REV(Seg s){
    swap(s.l, s.r);
    return s;
}

ll INF = 19191919;
struct LazySegTree{
    int n;
    vector<ll> lazy;
    vector<Seg> dat;

    //初期化
    LazySegTree(){}
    LazySegTree(int _n){
        n=1;
        while(n<_n) n*=2;
        dat = vector<Seg>(2*n-1);
        lazy = vector<ll>(2*n-1,INF);
    }

    void init(const vector<int> &w){
        int sz = w.size();
        rep(i,sz){
            int idx = hl.vid[i] + n-1;
            dat[idx] = Seg(w[i]);
        }
        for(int i=n-2; i>=0; --i) dat[i] = MERGE(dat[2*i+1],dat[2*i+2]);
    }

    void setLazy(int k, ll v){
        lazy[k] = v;

        dat[k].w = v*dat[k].sz;

        if(v>=0){
            dat[k].l = dat[k].r = dat[k].mx = dat[k].nb = dat[k].w;
        }
        else{
            dat[k].l = dat[k].r = dat[k].mx = 0;
            dat[k].nb = v;
        }
    }

    void push(int k, int l, int r){
        if(lazy[k]!=INF){
            setLazy(2*k+1,lazy[k]);
            setLazy(2*k+2,lazy[k]);
        }
        lazy[k]=INF;
    }

    void fix(int k, int l, int r){
        dat[k] = MERGE(dat[2*k+1],dat[2*k+2]);
    }

    Seg merge(Seg x, Seg y){
        return MERGE(x,y);
    }

    //内部的に投げられるクエリ
    void _update(int a, int b, ll x, int k, int l, int r){
        if(r<=a || b<=l) return;
        if(a<=l && r<=b){
            setLazy(k,x);
            return;
        }

        push(k,l,r);
        _update(a,b,x,2*k+1,l,(l+r)/2);
        _update(a,b,x,2*k+2,(l+r)/2,r);

        fix(k,l,r);
    }
    //[a,b)をxに更新
    void update(int a, int b, ll x){
        return _update(a,b,x,0,0,n);
    }

    //内部的に投げられるクエリ
    Seg _query(int a, int b, int k, int l, int r){
        if(r<=a || b<=l) return Seg();
        if(a<=l && r<=b) return dat[k];

        push(k,l,r);
        Seg vl=_query(a,b,2*k+1,l,(l+r)/2);
        Seg vr=_query(a,b,2*k+2,(l+r)/2,r);
        return merge(vl,vr);
    }
    //[a,b)
    Seg query(int a, int b){
        return _query(a,b,0,0,n);
    }
};

LazySegTree st;

void _UPDATE(int p, int c, int val){
    while(1){
        if(hl.head[p] == hl.head[c]){
            st.update(hl.vid[p], hl.vid[c]+1, val);
            break;
        }
        else{
            st.update(hl.vid[hl.head[c]], hl.vid[c]+1, val);
            c = hl.par[hl.head[c]];
        }
    }
}

void UPDATE(int u, int v, int val){
    int lca = hl.lca(u,v);
    _UPDATE(lca,u,val);
    _UPDATE(lca,v,val);
}

ll QUERY(int u, int v){
    int p = hl.lca(u,v);

    Seg sl,sr;

    // printf(" u -> p  %d %d\n",u,p);
    int c = u;
    while(1){
        if(hl.head[p] == hl.head[c]){
            Seg as = st.query(hl.vid[p], hl.vid[c]+1);
            as = REV(as);
            sl = MERGE(sl, as);

            // printf(" (%d %d)\n", p, c);
            // as.show();
            // sl.show();
            break;
        }
        else{
            Seg as = st.query(hl.vid[hl.head[c]], hl.vid[c]+1);
            as = REV(as);
            sl = MERGE(sl, as);

            // printf(" (%d %d)\n", hl.head[c], c);
            // as.show();
            // sl.show();
            c = hl.par[hl.head[c]];
        }
    }

    // printf(" v -> p  %d %d\n",v,p);
    c = v;
    while(1){
        if(hl.head[p] == hl.head[c]){
            Seg as = st.query(hl.vid[p]+1, hl.vid[c]+1);
            sr = MERGE(as, sr);

            // printf(" (%d %d)\n", p, c);
            // sr.show();
            break;
        }
        else{
            Seg as = st.query(hl.vid[hl.head[c]], hl.vid[c]+1);
            sr = MERGE(as, sr);

            // printf(" (%d %d)\n", hl.head[c], c);
            // sr.show();
            c = hl.par[hl.head[c]];
        }
    }

    Seg res = MERGE(sl,sr);
    // res.show();
    ll ans = res.mx;
    if(ans == 0) ans = res.nb;
    return ans;
}

int main(){
    int n,q;
    scanf(" %d %d", &n, &q);

    hl = HL_decomposition(n);
    st = LazySegTree(n);

    vector<int> w(n);
    rep(i,n) scanf(" %d", &w[i]);

    rep(i,n-1){
        int u,v;
        scanf(" %d %d", &u, &v);
        --u;
        --v;
        hl.add_edge(u,v);
    }

    hl.build();
    st.init(w);

    rep(i,q){
        int t,a,b,c;
        scanf(" %d %d %d %d", &t, &a, &b, &c);
        --a;
        --b;
        if(t==1) UPDATE(a,b,c);
        else printf("%lld\n", QUERY(a,b));
    }
    return 0;
}