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;
  class Fenwick {
    vector<long long> tree;
  
   public:
    Fenwick() {}
    Fenwick(int n, long long init) : tree(n + 5, 0) {}
    void add(int idx, long long val) {
      for (int i = idx; i < (int)tree.size(); i += i & -i) {
        tree[i] += val;
      }
    }
    long long get(int idx) {
      long long ans = 0;
      for (int i = idx; i > 0; i -= i & -i) {
        ans += tree[i];
      }
      return ans;
    }
    long long query(int l, int r) {
      long long ans = get(r) - get(l - 1);
      return ans;
    }
    void set(int idx, long long val) { add(idx, val - query(idx, idx)); }
  };
  template <typename T, class DS>
  class HLD {
    int vertexc;
    vector<int> *adj;
    vector<int> subtree_size;
    DS structure;
    DS aux;
    void build_sizes(int vertex, int parent) {
      subtree_size[vertex] = 1;
      for (int child : adj[vertex]) {
        if (child != parent) {
          build_sizes(child, vertex);
          subtree_size[vertex] += subtree_size[child];
        }
      }
    }
    int cur;
    vector<int> ord;
    vector<int> chain_root;
    vector<int> par;
    void build_hld(int vertex, int parent, int chain_source) {
      cur++;
      ord[vertex] = cur;
      chain_root[vertex] = chain_source;
      par[vertex] = parent;
      if (adj[vertex].size() > 1 || (vertex == 1 && adj[vertex].size() == 1)) {
        int big_child, big_size = -1;
        for (int child : adj[vertex]) {
          if ((child != parent) && (subtree_size[child] > big_size)) {
            big_child = child;
            big_size = subtree_size[child];
          }
        }
        build_hld(big_child, vertex, chain_source);
        for (int child : adj[vertex]) {
          if ((child != parent) && (child != big_child))
            build_hld(child, vertex, child);
        }
      }
    }
  
   public:
    HLD(int _vertexc) {
      vertexc = _vertexc;
      adj = new vector<int>[vertexc + 5];
    }
    void add_edge(int u, int v) {
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    void build(T initial) {
      subtree_size = vector<int>(vertexc + 5);
      ord = vector<int>(vertexc + 5);
      chain_root = vector<int>(vertexc + 5);
      par = vector<int>(vertexc + 5);
      cur = 0;
      build_sizes(1, -1);
      build_hld(1, -1, 1);
      structure = DS(vertexc + 5, initial);
      aux = DS(50, initial);
    }
    void set(int vertex, int value) { structure.set(ord[vertex], value); }
    T query_path(int u, int v) {
      int cur_id = 0;
      while (chain_root[u] != chain_root[v]) {
        if (ord[u] > ord[v]) {
          cur_id++;
          aux.set(cur_id, structure.query(ord[chain_root[u]], ord[u]));
          u = par[chain_root[u]];
        } else {
          cur_id++;
          aux.set(cur_id, structure.query(ord[chain_root[v]], ord[v]));
          v = par[chain_root[v]];
        }
      }
      cur_id++;
      aux.set(cur_id, structure.query(min(ord[u], ord[v]), max(ord[u], ord[v])));
      return aux.query(1, cur_id);
    }
    void print() {
      for (int i = 1; i <= vertexc; i++)
        cout << i << ' ' << ord[i] << ' ' << chain_root[i] << ' ' << par[i]
             << endl;
    }
  };
  const int MAX_N = 1e5 + 5;
  long long arr[MAX_N];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, qc;
    cin >> n >> qc;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
      arr[i] = abs(arr[i]);
    }
    HLD<long long, Fenwick> hld(n + 5);
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      hld.add_edge(u, v);
    }
    hld.build(1);
    for (int i = 1; i <= n; i++) {
      hld.set(i, arr[i]);
    }
    for (int i = 0; i < qc; i++) {
      int type;
      cin >> type;
      if (type == 1) {
        int u, c;
        cin >> u >> c;
        arr[u] = abs(c);
        hld.set(u, arr[u]);
      } else {
        int u, v;
        cin >> u >> v;
        long long ans = hld.query_path(u, v);
        ans *= 2;
        ans -= arr[u];
        ans -= arr[v];
        cout << ans << '\n';
      }
    }
  }
  