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;
  vector<int> e[100010];
  int d[100010], in[100010], out[100010], c;
  int par[18][100010];
  void dfs(int v, int p = -1) {
    in[v] = ++c;
    for (int i = 0; i < e[v].size(); i++) {
      int w = e[v][i];
      if (p == w) continue;
      par[0][w] = v;
      d[w] = d[v] + 1;
      dfs(w, v);
    }
    out[v] = ++c;
  }
  long long b[200010];
  void add(int p, int v) {
    while (p < 200010) {
      b[p] += v;
      p += p & -p;
    }
  }
  long long sum(int p) {
    long long res = 0;
    while (p) {
      res += b[p];
      p -= p & -p;
    }
    return res;
  }
  int lca(int u, int v) {
    if (d[u] > d[v]) return lca(v, u);
    int dd = d[v] - d[u];
    for (int i = 0; i < 17; i++) {
      if (dd >> i & 1) {
        v = par[i][v];
      }
    }
    if (u == v) return u;
    for (int i = 16; i >= 0; i--) {
      if (par[i][u] != par[i][v]) {
        u = par[i][u];
        v = par[i][v];
      }
    }
    return par[0][u];
  }
  int a[100010];
  int main() {
    int n, q;
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      if (a[i] < 0) a[i] = -a[i];
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      --u;
      --v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    dfs(0);
    for (int i = 0; i < n; i++) {
      add(in[i], a[i]);
      add(out[i], -a[i]);
    }
    for (int i = 0; i < 16; i++) {
      for (int j = 0; j < n; j++) {
        par[i + 1][j] = par[i][par[i][j]];
      }
    }
    while (q--) {
      int ty, u, v;
      scanf("%d%d%d", &ty, &u, &v);
      --u;
      if (ty == 1) {
        add(in[u], -a[u]);
        add(out[u], a[u]);
        a[u] = v > 0 ? v : -v;
        add(in[u], a[u]);
        add(out[u], -a[u]);
      } else {
        --v;
        int w = lca(u, v);
        long long ans = 0;
        if (w == u) {
          ans = (sum(in[v]) - sum(in[u])) * 2 - a[v] + a[u];
        } else if (w == v) {
          ans = (sum(in[u]) - sum(in[v])) * 2 - a[u] + a[v];
        } else {
          ans = 2 * (sum(in[v]) + sum(in[u])) - 4 * sum(in[w]) + 2 * a[w] - a[u] -
                a[v];
        }
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  