Chanek Jones is back, helping his long-lost relative Indiana Jones, to find a secret treasure in a maze buried below a desert full of illusions.

The map of the labyrinth forms a tree with n rooms numbered from 1 to n and n - 1 tunnels connecting them such that it is possible to travel between each pair of rooms through several tunnels.

The i-th room (1 ≤ i ≤ n) has a_i illusion rate. To go from the x-th room to the y-th room, there must exist a tunnel between x and y, and it takes max(|a_x + a_y|, |a_x - a_y|) energy. |z| denotes the absolute value of z.

To prevent grave robbers, the maze can change the illusion rate of any room in it. Chanek and Indiana would ask q queries.

There are two types of queries to be done:

  * 1\ u\ c — The illusion rate of the x-th room is changed to c (1 ≤ u ≤ n, 0 ≤ |c| ≤ 10^9). 
  * 2\ u\ v — Chanek and Indiana ask you the minimum sum of energy needed to take the secret treasure at room v if they are initially at room u (1 ≤ u, v ≤ n). 



Help them, so you can get a portion of the treasure!

Input

The first line contains two integers n and q (2 ≤ n ≤ 10^5, 1 ≤ q ≤ 10^5) — the number of rooms in the maze and the number of queries.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ |a_i| ≤ 10^9) — inital illusion rate of each room.

The i-th of the next n-1 lines contains two integers s_i and t_i (1 ≤ s_i, t_i ≤ n), meaning there is a tunnel connecting s_i-th room and t_i-th room. The given edges form a tree.

The next q lines contain the query as described. The given queries are valid.

Output

For each type 2 query, output a line containing an integer — the minimum sum of energy needed for Chanek and Indiana to take the secret treasure.

Example

Input


6 4
10 -9 2 -1 4 -6
1 5
5 4
5 6
6 2
6 3
2 1 2
1 1 -3
2 1 2
2 3 3


Output


39
32
0

Note

<image>

In the first query, their movement from the 1-st to the 2-nd room is as follows.

  * 1 → 5 — takes max(|10 + 4|, |10 - 4|) = 14 energy. 
  * 5 → 6 — takes max(|4 + (-6)|, |4 - (-6)|) = 10 energy. 
  * 6 → 2 — takes max(|-6 + (-9)|, |-6 - (-9)|) = 15 energy. 

In total, it takes 39 energy.

In the second query, the illusion rate of the 1-st room changes from 10 to -3.

In the third query, their movement from the 1-st to the 2-nd room is as follows.

  * 1 → 5 — takes max(|-3 + 4|, |-3 - 4|) = 7 energy. 
  * 5 → 6 — takes max(|4 + (-6)|, |4 - (-6)|) = 10 energy. 
  * 6 → 2 — takes max(|-6 + (-9)|, |-6 - (-9)|) = 15 energy. 



Now, it takes 32 energy.
#include <bits/stdc++.h>                                           
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;

#define int long long
//#pragma GCC optimize("Ofast")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
 
 
#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);
#define fastio ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define all(x) x.begin(), x.end()
#define sz(s) (int)s.size()
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define s second
#define f first
 
 
 
 
typedef pair < long long, long long > pll;    
typedef pair < int, int > pii;  
typedef unsigned long long ull;         
typedef vector < pii > vpii;                                    
typedef vector < int > vi;
typedef long double ldb;  
typedef long long ll;  
typedef double db;
 
typedef tree < int, null_type, less < int >, rb_tree_tag, tree_order_statistics_node_update > ordered_set;
 
const int inf = 1e9, maxn = 2e5 + 48, mod = 1e9 + 7, N = 1e5 + 12;
const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 400;
const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);
const db eps = 1e-12, pi = acos(-1);
const ll INF = 1e18;


int n, q, a[N];
vector<int> g[N];
int tin[N], tout[N], timer, up[N][20];

void dfs(int v, int par = 0){
    tin[v] = ++timer;
    up[v][0] = par;
    for(int i = 1; i < 20; i++)
        up[v][i] = up[up[v][i - 1]][i - 1];
    
    for(auto to : g[v]){
        if(to != par){
            dfs(to, v);    
        }
    }

    tout[v] = timer;
}
bool is_par(int u, int v){
    return (tin[u] <= tin[v] && tout[v] <= tout[u]);    
}

int lca(int a, int b){
    if(is_par(a, b)) return a;
    if(is_par(b, a)) return a;
    for(int i = 19; i >= 0; i--)
        if(!is_par(up[a][i], b))
            a = up[a][i];
    return up[a][0];
}
int t[N];
         
void upd(int pos, int val){
    for(; pos < N; pos |= (pos + 1))
        t[pos] += val;
}

void upd(int l, int r, int x){
    upd(l, x);
    upd(r + 1, -x);
}

int get(int r){
    int res = 0;
    for(; r >= 0; r = (r & (r + 1)) - 1)
        res += t[r];
    return res;
}

main () {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> q;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
        a[i] = abs(a[i]);
    }  
    for(int i = 1; i < n; i++){
        int u, v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    dfs(1);
    
    for(int i = 1; i <= n; i++)
        upd(tin[i], tout[i], a[i]);

    while(q--){
        int type;
        int u, v, x;
        cin >> type;
        if(type == 2){
            cin >> u >> v;
            if(u == v){
                cout << "0\n";
                continue;
            }                     
            int l = lca(u, v);
            cout << 2 * (get(tin[u]) + get(tin[v]) - get(tin[l]) - get(tin[up[l][0]])) - a[u] - a[v] << "\n";
        }else{
            cin >> v >> x;
            x = abs(x);
            upd(tin[v], tout[v], x - a[v]);
            a[v] = x;
        }
    }  

}