description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int Nmax = 1e5 + 10;
  int n, m;
  vector<int> e[Nmax];
  int par[Nmax][20];
  int in[Nmax], out[Nmax], ti;
  void dfs(int u, int p) {
    in[u] = ++ti;
    par[u][0] = p;
    for (int i = 1; i < 20; ++i) par[u][i] = par[par[u][i - 1]][i - 1];
    for (auto v : e[u]) {
      if (v == p) continue;
      dfs(v, u);
    }
    out[u] = ti;
  }
  long long BIT[Nmax];
  void add(int id, int delta) {
    for (; id <= n; id += id & (-id)) BIT[id] += delta;
  }
  long long get(int id) {
    long long ans = 0;
    for (; id; id -= id & (-id)) ans += BIT[id];
    return ans;
  }
  int check(int u, int v) { return in[u] <= in[v] && out[v] <= out[u]; }
  int LCA(int u, int v) {
    if (check(u, v)) return u;
    for (int i = 19; i >= 0; --i)
      if (par[u][i] && !check(par[u][i], v)) u = par[u][i];
    return par[u][0];
  }
  int a[Nmax];
  int main() {
    ios ::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    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;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; ++i) add(in[i], a[i]), add(out[i] + 1, -a[i]);
    while (m--) {
      int type;
      cin >> type;
      if (type == 1) {
        int u, c;
        cin >> u >> c;
        c = abs(c);
        add(in[u], c - a[u]);
        add(out[u] + 1, a[u] - c);
        a[u] = c;
      } else {
        int u, v;
        cin >> u >> v;
        int w = LCA(u, v);
        cout << (get(in[u]) + get(in[v]) - 2 * get(in[w]) + a[w]) * 2 - a[u] -
                    a[v]
             << '\n';
      }
    }
  }
  