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<cstdio>
#include<cstring>
#include<map>
#include<queue>
#include<vector>
#include<string>
#include<iostream>
#include<algorithm>

using namespace std;

#define MP make_pair
#define PB push_back
#define foreach(e,x) for(__typeof(x.begin()) e=x.begin(); e!=x.end(); ++e)

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> mapII;

const int MAX_N = 400000 + 10;
const int INF = 2000000 + 10;
const double EPS = 1e-8;

struct INFO
{
	LL maxL, maxR, ans, sum;
	INFO() {
		maxL = maxR = ans = - INF;
		sum = 0;
	}
};

INFO operator + (const INFO &a, const INFO &b) 
{
	INFO ret;
	ret.maxL = max(a.maxL, a.sum + b.maxL);
	ret.maxR = max(b.maxR, b.sum + a.maxR);
	ret.ans = max(max(a.ans, b.ans), a.maxR + b.maxL);
	ret.sum = a.sum + b.sum;
	return ret;
}

struct Tree
{
	int l, r, cover, size;
	INFO info;
	Tree *ch[2];
	Tree() {}
	Tree(int _l, int _r, int *sqn) {
		l = _l, r = _r;
		cover = INF;
		if (l + 1 == r) {
			info.sum = sqn[l];
			info.ans = sqn[l];
			info.maxL = info.maxR = sqn[l];
			size = 1;
			return;
		}
		int mid = (l + r) >> 1;
		ch[0] = new Tree(l, mid, sqn);
		ch[1] = new Tree(mid, r, sqn);
		info = ch[0]->info + ch[1]->info;
		size = ch[0]->size + ch[1]->size;
	}

	void appCover(int x) {
		cover = x;
		info.sum = x * size;
		if (x >= 0) {
			info.maxL = info.maxR = info.ans = x * size;
		} else {
			info.maxL = info.maxR = info.ans = x;
		}
	}

	void relax() {
		if (cover == INF) return;
		ch[0]->appCover(cover);
		ch[1]->appCover(cover);
		cover = INF;
	}

	INFO query(int _l, int _r) {
		if (_l <= l && _r >= r) 
			return info;
		if (_r <= l || _l >= r) return INFO();
		relax();
		return ch[0]->query(_l, _r) + ch[1]->query(_l, _r);
	}

	void modify(int _l, int _r, int c) {
		if (_l <= l && _r >= r) {
			appCover(c);
			return;
		}
		if (_r <= l || _l >= r) 
			return;
		relax();
		ch[0]->modify(_l, _r, c);
		ch[1]->modify(_l, _r, c);
		info = ch[0]->info + ch[1]->info;
	}
};

int N, M, val[MAX_N];
vector<int> adj[MAX_N];
int father[MAX_N], size[MAX_N], depth[MAX_N];
int bfsOrd[MAX_N], pathId[MAX_N], ordInPath[MAX_N], sqn[MAX_N];
Tree *root[MAX_N];

void doBfs(int s)
{
	int *que = bfsOrd;
	int qh = 0, qt = 0;
	father[s] = -1; depth[s] = 0;

	for(que[qt ++] = s; qh < qt; ) {
		int u = que[qh ++];
		foreach(iter, adj[u]) {
			int v = *iter;
			if (v == father[u])
				continue;
			father[v] = u;
			depth[v] = depth[u] + 1;
			que[qt ++] = v;
		}
	}
}

void doSplit()
{
	for(int i = N - 1; i >= 0; -- i) {
		int u = bfsOrd[i];
		size[u] = 1;
		foreach(iter, adj[u]) {
			int v = *iter;
			if (v == father[u]) continue;
			size[u] += size[v];
		}
	}

	memset(pathId, -1, sizeof pathId);
	for(int i = 0; i < N; ++ i) {
		int top = bfsOrd[i];
		if (pathId[top] != -1)
			continue;
		int cnt = 0;
		for(int u = top; u != -1; ) {
			sqn[cnt] = val[u];
			ordInPath[u] = cnt;
			pathId[u] = top;
			++ cnt;

			int next = -1;
			foreach(iter, adj[u]) {
				int v = *iter;
				if (v == father[u])
					continue;
				if (next < 0 || size[next] < size[v])
					next = v;
			}
			u = next;
		}
		root[top] = new Tree(0, cnt, sqn);
	}
}

void modify(int u, int v, int c)
{
	int top;
	for( ; pathId[u] != pathId[v]; ) {
		if (depth[pathId[u]] > depth[pathId[v]]) {
			top = pathId[u];
			root[top]->modify(0, ordInPath[u] + 1, c);
			u = father[top];
		} else {
			top = pathId[v];
			root[top]->modify(0, ordInPath[v] + 1, c);
			v = father[top];
		}
	}
	top = pathId[u];
	int l = min(ordInPath[v], ordInPath[u]);
	int r = max(ordInPath[v], ordInPath[u]);
	root[top]->modify(l, r + 1, c);
}

int query(int u, int v)
{
	int top;
	INFO infou, infov;
	for( ; pathId[u] != pathId[v]; ) {
		if (depth[pathId[u]] > depth[pathId[v]]) {
			top = pathId[u];
			infou = root[top]->query(0, ordInPath[u] + 1) + infou;
			u = father[top];
		} else {
			top = pathId[v];
			infov = root[top]->query(0, ordInPath[v] + 1) + infov;
			v = father[top];
		}
	}
	
	top = pathId[u];
	if (ordInPath[u] < ordInPath[v]) {
		swap(infou.maxL, infou.maxR);
		infou = infou + root[top]->query(ordInPath[u], ordInPath[v] + 1);
		infou = infou + infov;
		return (int)infou.ans;
	} else {
		swap(infov.maxL, infov.maxR);
		infov = infov + root[top]->query(ordInPath[v], ordInPath[u] + 1);
		infov = infov + infou;
		return (int)infov.ans;
	}
}

void solve()
{
	int u, v;
	scanf("%d%d", &N, &M);
	for(int i = 0; i < N; ++ i) {
		scanf("%d", &val[i]);
		adj[i].clear();
	}
	for(int i = 0; i < N - 1; ++ i) {
		scanf("%d%d", &u, &v);
		-- u, -- v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}

	doBfs(0);
	doSplit();

	int t, c;
	for(int i = 0; i < M; ++ i) {
		scanf("%d%d%d%d", &t, &u, &v, &c);
		-- u, -- v;
		if (t == 1) {
			modify(u, v, c);
		} else {
			printf("%d\n", query(u, v));
		}
	}
}

int main()
{
	solve();
	return 0;
}