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 + 1;
  vector<vector<int>> e(N, vector<int>());
  struct LCA {
    vector<int> tin, tout;
    vector<int> depth;
    vector<vector<int>> lca;
    int time = 1;
    LCA(int n) {
      tin.resize(n + 1);
      tout.resize(n + 1);
      depth.resize(n + 1);
      lca.resize(n + 1);
      int h = (int)log2(n);
      for (int i = 0; i <= n; i++) lca[i] = vector<int>(h + 1);
      dfs(1, 0);
    }
    void dfs(int u, int p) {
      tin[u] = time++;
      for (int v : e[u]) {
        if (v != p) {
          lca[v][0] = u;
          for (int lg = 1; lg < (int)lca[1].size(); lg++) {
            lca[v][lg] = lca[lca[v][lg - 1]][lg - 1];
          }
          depth[v] = depth[u] + 1;
          dfs(v, u);
        }
      }
      tout[u] = time++;
    }
    int query(int u, int v) {
      if (depth[u] > depth[v]) swap(u, v);
      for (int lg = 0; lg < (int)lca[1].size(); lg++) {
        if ((1 << lg) & (depth[v] - depth[u])) {
          v = lca[v][lg];
        }
      }
      assert(depth[u] == depth[v]);
      if (u == v) {
        return u;
      }
      for (int lg = (int)lca[1].size() - 1; lg >= 0; lg--) {
        if (lca[u][lg] != lca[v][lg]) {
          u = lca[u][lg];
          v = lca[v][lg];
        }
      }
      assert(lca[u][0] == lca[v][0]);
      return lca[u][0];
    }
  };
  struct fenwick {
    vector<int64_t> fen;
    fenwick(int n) { fen.resize(n); }
    void upd(int i, int val) {
      while (i < (int)fen.size()) {
        fen[i] += val;
        i += i & -i;
      }
    }
    int64_t get(int i) {
      int64_t r = 0;
      while (i > 0) {
        r += fen[i];
        i -= i & -i;
      }
      return r;
    }
  };
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, q;
    cin >> n >> q;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
      e[i].clear();
    }
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      cin >> u >> v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    fenwick *fen = new fenwick(2 * n + 3);
    LCA *lca = new LCA(n);
    for (int i = 1; i <= n; i++) {
      fen->upd(lca->tin[i], a[i]);
      fen->upd(lca->tout[i] + 1, -a[i]);
    }
    while (q--) {
      int op;
      cin >> op;
      if (op == 1) {
        int u, c;
        cin >> u >> c;
        c = abs(c);
        int inc = c - a[u];
        fen->upd(lca->tin[u], inc);
        fen->upd(lca->tout[u] + 1, -inc);
        a[u] = c;
      } else {
        int u, v;
        cin >> u >> v;
        int ancestor = lca->query(u, v);
        cout << 2LL * (fen->get(lca->tin[u]) + fen->get(lca->tin[v]) -
                       2 * fen->get(lca->tin[ancestor]) + a[ancestor]) -
                    a[u] - a[v]
             << "\n";
      }
    }
    return 0;
  }
  