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;

struct RURMS {
	static const ll INF = 1e17;
	struct t1 {
		ll left, right, all, sum, size;
		t1() : left(-INF), right(-INF), all(-INF), sum(0), size(0) {}
		t1(ll val) : left(val), right(val), all(val), sum(val), size(1) {}
		t1(ll l, ll r, ll a, ll s, ll si) : left(l), right(r), all(a), sum(s), size(si) {}
		bool operator==(const t1& r) const {
			return left == r.left && right == r.right && all == r.all && sum == r.sum && size == r.size;
		}
	};
	using t2 = ll;
	static t1 id1() { return t1(); }
	static t2 id2() { return -INF; }
	static t1 op1(const t1& l, const t1 & r) {
		return t1(max(l.left, l.sum + r.left), max(r.right, r.sum + l.right), max({ l.all, r.all, l.right + r.left }), l.sum + r.sum, l.size + r.size);
	}
	static t1 op2(const t1& l, const t2 & r) {
		if (r == id2()) return l;
		if (r > 0) return t1(r * l.size, r * l.size, r * l.size, r * l.size, l.size);
		return t1(r, r, r, r * l.size, l.size);
	}
	static t2 op3(const t2& l, const t2 & r) {
		return r == id2() ? l : r;
	}
};

template <typename M>
class LazySegmentTree {
	using T1 = typename M::t1;
	using T2 = typename M::t2;
	const int h, n;
	vector<T1> data;
	vector<T2> lazy;
	void push(int node) {
		if (lazy[node] == M::id2()) return;
		if (node < n) {
			lazy[node * 2] = M::op3(lazy[node * 2], lazy[node]);
			lazy[node * 2 + 1] = M::op3(lazy[node * 2 + 1], lazy[node]);
		}
		data[node] = M::op2(data[node], lazy[node]);
		lazy[node] = M::id2();
	}
	void update(int node) {
		data[node] = M::op1(M::op2(data[node * 2], lazy[node * 2]), M::op2(data[node * 2 + 1], lazy[node * 2 + 1]));
	}
public:
	LazySegmentTree(int n_)
		: h(ceil(log2(n_))), n(1 << h), data(n * 2, M::id1()), lazy(n * 2, M::id2()) {}
	LazySegmentTree(int n_, T1 v1, T2 v2)
		: h(ceil(log2(n_))), n(1 << h), data(n * 2, v1), lazy(n * 2, v2) {}
	LazySegmentTree(const vector<T1>& data_)
		: h(ceil(log2(data_.size()))), n(1 << h), data(n * 2, M::id1()), lazy(n * 2, M::id2()) {
		init(data_);
	}
	void init() {
		for (int i = n - 1; i >= 1; i--) data[i] = M::op1(data[i * 2], data[i * 2 + 1]);
	}
	void init(const vector<T1>& data_) {
		for (int i = 0; i < (int)data_.size(); i++) data[i + n] = data_[i];
		init();
	}
	void update(int l, int r, T2 val) {
		l += n, r += n;
		for (int i = h; i > 0; i--) push(l >> i), push(r >> i);
		int tl = l, tr = r;
		r++;
		while (l < r) {
			if (l & 1) lazy[l] = M::op3(lazy[l], val), l++;
			if (r & 1) r--, lazy[r] = M::op3(lazy[r], val);
			l >>= 1; r >>= 1;
		}
		while (tl >>= 1, tr >>= 1, tl) {
			if (lazy[tl] == M::id2()) update(tl);
			if (lazy[tr] == M::id2()) update(tr);
		}
	}
	T1 find(int l, int r) {
		l += n, r += n;
		for (int i = h; i > 0; i--) push(l >> i), push(r >> i);
		r++;
		T1 res1 = M::id1(), res2 = M::id1();
		while (l < r) {
			if (l & 1) res1 = M::op1(res1, M::op2(data[l], lazy[l])), l++;
			if (r & 1) r--, res2 = M::op1(M::op2(data[r], lazy[r]), res2);
			l >>= 1; r >>= 1;
		}
		return M::op1(res1, res2);
	}
};

class HLD {
	vector<vector<int>> G;
	vector<int> vid, head, heavy, par, dep;
	int dfs(int v, int prev) {
		par[v] = prev;
		int cnt = 1, ma = 0;
		for (int to : G[v]) if (to != prev) {
			dep[to] = dep[v] + 1;
			int c = dfs(to, v);
			cnt += c;
			if (ma < c) {
				ma = c;
				heavy[v] = to;
			}
		}
		return cnt;
	}
public:
	HLD(int n) : G(n), vid(n, -1), head(n), heavy(n, -1), par(n), dep(n) {}
	void add(int u, int v) {
		G[u].push_back(v);
		G[v].push_back(u);
	}
	void build() {
		dfs(0, -1);
		int k = 0;
		queue<int> q;
		q.push(0);
		while (!q.empty()) {
			int h = q.front(); q.pop();
			for (int i = h; i != -1; i = heavy[i]) {
				vid[i] = k++;
				head[i] = h;
				for (int j : G[i]) if (j != par[i] && j != heavy[i]) q.push(j);
			}
		}
	}
	int operator[](int v) {
		return vid[v];
	}
	void for_each(int u, int v, function<void(int, int)> f) {
		while (true) {
			if (vid[u] > vid[v]) swap(u, v);
			f(max(vid[head[v]], vid[u]), vid[v]);
			if (head[u] == head[v]) break;
			v = par[head[v]];
		}
	}
	void for_each_directed(int u, int v, function<void(int, int, int)> f) {
		if (vid[u] > vid[v]) {
			f(max(vid[head[u]], vid[v]), vid[u], 1);
			if (head[u] != head[v]) for_each_directed(par[head[u]], v, f);
		}
		else {
			f(max(vid[head[v]], vid[u]), vid[v], 0);
			if (head[u] != head[v]) for_each_directed(u, par[head[v]], f);
		}
	}
};

int main()
{
	ios::sync_with_stdio(false), cin.tie(0);
	int N, Q;
	cin >> N >> Q;
	vector<ll> A(N);
	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}
	HLD hld(N);
	for (int i = 0; i < N - 1; i++) {
		int u, v;
		cin >> u >> v; u--, v--;
		hld.add(u, v);
	}
	hld.build();
	vector<RURMS::t1> base(N);
	for (int i = 0; i < N; i++) {
		base[hld[i]] = RURMS::t1(A[i]);
	}
	LazySegmentTree<RURMS> lst(base);
	while (Q--) {
		int t, a, b, c;
		cin >> t >> a >> b >> c; a--, b--;
		if (t == 1) {
			hld.for_each(a, b, [&](int l, int r) {
				lst.update(l, r, c);
			});
		}
		else {
			RURMS::t1 res1, res2;
			hld.for_each_directed(a, b, [&](int l, int r, int b) {
				if (l > r) swap(l, r);
				if (b) {
					auto t = lst.find(l, r); swap(t.left, t.right);
					res1 = RURMS::op1(res1, t);
				}
				else {
					res2 = RURMS::op1(lst.find(l, r), res2);
				}
			});
			printf("%d\n", (int)RURMS::op1(res1, res2).all);
		}
	}
	return 0;
}