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 N = 1e5 + 50;
  int n, q, dfn[N], rr[N], cnt, f[N][20], d[N], a[N];
  long long bit[N];
  vector<int> v[N];
  void dfs(int x, int ff) {
    dfn[x] = ++cnt;
    f[x][0] = ff;
    d[x] = d[ff] + 1;
    for (int i = 0; i < 18; i++) f[x][i + 1] = f[f[x][i]][i];
    for (int i = 0; i < v[x].size(); i++) {
      int y = v[x][i];
      if (y == ff) continue;
      dfs(y, x);
    }
    rr[x] = cnt;
  }
  int lca(int x, int y) {
    if (d[x] < d[y]) swap(x, y);
    for (int i = 18; ~i; i--)
      if (d[f[x][i]] >= d[y]) x = f[x][i];
    if (x == y) return x;
    for (int i = 18; ~i; i--)
      if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
    return f[x][0];
  }
  void add(int x, int y) {
    for (; x <= n; x += x & -x) bit[x] += y;
  }
  long long ask(int x) {
    long long z = 0;
    for (; x; x -= x & -x) z += bit[x];
    return z;
  }
  int main() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] = abs(a[i]);
    for (int i = 1, x, y; i < n; i++) {
      scanf("%d%d", &x, &y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++) add(dfn[i], a[i]), add(rr[i] + 1, -a[i]);
    for (int i = 1, op, x, y; i <= q; i++) {
      scanf("%d%d%d", &op, &x, &y);
      if (op == 1) {
        add(dfn[x], -a[x]);
        add(rr[x] + 1, a[x]);
        add(dfn[x], a[x] = abs(y));
        add(rr[x] + 1, -a[x]);
      } else {
        int z = lca(x, y), zz = f[z][0];
        printf("%lld\n",
               2 * (ask(dfn[x]) + ask(dfn[y]) - ask(dfn[z]) - ask(dfn[zz])) -
                   a[x] - a[y]);
      }
    }
    return 0;
  }
  