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;

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);
		}
	}
};

const ll INF = 1e17;

struct RURMQ {
	struct type {
		ll left, right, all, sum, size;
		type() : left(-INF), right(-INF), all(-INF), sum(0), size(0) {}
		type(ll val) : left(val), right(val), all(val), sum(val), size(1) {}
		type(ll l, ll r, ll a, ll s, ll si) : left(l), right(r), all(a), sum(s), size(si) {}
		bool operator==(const type& r) const {
			return left == r.left && right == r.right && all == r.all && sum == r.sum && size == r.size;
		}
	};
	static type id1() { return type(); }
	static type id2() { return type(-INF, -INF, -INF, -INF, -1); }
	static type op1(const type& l, const type & r) {
		return type(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 type op2(const type& l, const type & r) {
		if (r == id2()) return l;
		if (l.size < 0) return r;
		ll val = r.all;
		if (val > 0) {
			return type(val * l.size, val * l.size, val * l.size, val * l.size, l.size);
		}
		else {
			return type(val, val, val, val * l.size, l.size);
		}
	}
};

template <typename M>
class lSegmentTree {
	using T = typename M::type;
	const int n;
	vector<T> data, lazy;
	int size(int n) {
		int res = 1;
		while (res < n) res <<= 1;
		return res;
	}
	void push(int node) {
		if (lazy[node] == M::id2()) return;
		if (node < n) {
			lazy[node * 2] = M::op2(lazy[node * 2], lazy[node]);
			lazy[node * 2 + 1] = M::op2(lazy[node * 2 + 1], lazy[node]);
		}
		data[node] = M::op2(data[node], lazy[node]);
		lazy[node] = M::id2();
	}
	void suc(int l, int r, int node, int lb, int ub, T val) {
		if (ub <= l || r <= lb) return;
		push(node);
		if (l <= lb && ub <= r) {
			lazy[node] = val;
			return;
		}
		int c = (lb + ub) / 2;
		suc(l, r, node * 2, lb, c, val);
		suc(l, r, node * 2 + 1, c, ub, val);
		data[node] = M::op1(M::op2(data[node * 2], lazy[node * 2]), M::op2(data[node * 2 + 1], lazy[node * 2 + 1]));
	}
	T sub(int l, int r, int node, int lb, int ub) {
		if (ub <= l || r <= lb) return M::id1();
		push(node);
		if (l <= lb && ub <= r) {
			return data[node];
		}
		int c = (lb + ub) / 2;
		return M::op1(sub(l, r, node * 2, lb, c), sub(l, r, node * 2 + 1, c, ub));
	}
public:
	lSegmentTree(int n_) : n(size(n_)), data(n * 2, M::id1()), lazy(n * 2, M::id2()) {}
	void init(const vector<T>& data_) {
		for (int i = 0; i < (int)data_.size(); i++)
			data[i + n] = data_[i];
		for (int i = n - 1; i >= 0; i--)
			data[i] = M::op1(data[i * 2], data[i * 2 + 1]);
		fill(lazy.begin(), lazy.end(), M::id2());
	}
	void update(int l, int r, T val) {
		suc(l, r + 1, 1, 0, n, val);
	}
	T find(int l, int r) {
		return sub(l, r + 1, 1, 0, n);
	}
};

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<RURMQ::type> base(N);
	for (int i = 0; i < N; i++) {
		base[hld[i]] = RURMQ::type(A[i]);
	}
	lSegmentTree<RURMQ> lst(N);
	lst.init(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, RURMQ::type(c, c, c, c, -1));
			});
		}
		else {
			RURMQ::type 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 = RURMQ::op1(res1, t);
				}
				else {
					res2 = RURMQ::op1(lst.find(l, r), res2);
				}
			});
			printf("%d\n", (int)RURMQ::op1(res1, res2).all);
		}
	}
	return 0;
}