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 + 5;
  struct segtree {
    int n;
    vector<long long> t;
    void init(int n, vector<long long>& a) {
      this->n = n;
      t.resize(n * 4);
      build(1, 0, n - 1, a);
    }
    inline void Merge(long long& node, long long& l, long long& r) {
      node = l + r;
    }
    void build(int node, int l, int r, vector<long long>& a) {
      if (l == r) {
        t[node] = a[l];
        return;
      }
      int mid = (l + r) / 2;
      build(node * 2, l, mid, a);
      build(node * 2 + 1, mid + 1, r, a);
      Merge(t[node], t[node * 2], t[node * 2 + 1]);
    }
    void update(int node, int l, int r, int i, int j, long long add) {
      if (l > j || r < i) {
        return;
      }
      if (l >= i && r <= j) {
        t[node] = add;
        return;
      }
      int mid = (l + r) / 2;
      update(node * 2, l, mid, i, j, add);
      update(node * 2 + 1, mid + 1, r, i, j, add);
      Merge(t[node], t[node * 2], t[node * 2 + 1]);
    }
    void update(int l, int r, int add) { update(1, 0, n - 1, l, r, add); }
    long long query(int node, int l, int r, int i, int j) {
      if (l > j || r < i) {
        return 0;
      }
      if (l >= i && r <= j) {
        return t[node];
      }
      int mid = (l + r) / 2;
      long long x = query(node * 2, l, mid, i, j);
      long long y = query(node * 2 + 1, mid + 1, r, i, j);
      return x + y;
    }
    long long query(int l, int r) {
      long long ret = query(1, 0, n - 1, l, r);
      return ret;
    }
  };
  struct HLD {
    vector<int> parent, depth, heavy, head, pos;
    int cur_pos;
    segtree seg;
    int dfs(int v, vector<vector<int>> const& adj) {
      int size = 1;
      int max_c_size = 0;
      for (int c : adj[v]) {
        if (c != parent[v]) {
          parent[c] = v, depth[c] = depth[v] + 1;
          int c_size = dfs(c, adj);
          size += c_size;
          if (c_size > max_c_size) max_c_size = c_size, heavy[v] = c;
        }
      }
      return size;
    }
    void decompose(int v, int h, vector<vector<int>> const& adj) {
      head[v] = h, pos[v] = cur_pos++;
      if (heavy[v] != -1) decompose(heavy[v], h, adj);
      for (int c : adj[v]) {
        if (c != parent[v] && c != heavy[v]) decompose(c, c, adj);
      }
    }
    void init(vector<vector<int>> const& adj, vector<long long>& a) {
      int n = adj.size();
      parent = vector<int>(n);
      depth = vector<int>(n);
      heavy = vector<int>(n, -1);
      head = vector<int>(n);
      pos = vector<int>(n);
      cur_pos = 0;
      dfs(0, adj);
      decompose(0, 0, adj);
      vector<long long> tmp(n);
      for (int i = 0; i < n; i++) {
        tmp[pos[i]] = a[i];
      }
      seg.init(n, tmp);
    }
    long long query(int a, int b) {
      long long res = 0;
      for (; head[a] != head[b]; b = parent[head[b]]) {
        if (depth[head[a]] > depth[head[b]]) swap(a, b);
        long long cur_heavy_path_max = seg.query(pos[head[b]], pos[b]);
        res += cur_heavy_path_max;
      }
      if (depth[a] > depth[b]) swap(a, b);
      long long last_heavy_path_max = seg.query(pos[a], pos[b]);
      res += last_heavy_path_max;
      return res;
    }
    void update(int a, int b, int x) {
      for (; head[a] != head[b]; b = parent[head[b]]) {
        if (depth[head[a]] > depth[head[b]]) swap(a, b);
        seg.update(pos[head[b]], pos[b], x);
      }
      if (depth[a] > depth[b]) swap(a, b);
      seg.update(pos[a], pos[b], x);
    }
  };
  int n, q;
  vector<vector<int>> adj;
  vector<long long> a;
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> q;
    a.resize(n);
    adj.resize(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    HLD h;
    h.init(adj, a);
    while (q--) {
      int t, u, v;
      cin >> t >> u >> v;
      if (t == 1) {
        v = abs(v);
        u--;
        h.update(u, u, v);
        a[u] = v;
      } else {
        u--;
        v--;
        if (u == v) {
          cout << "0\n";
          continue;
        }
        long long ans = h.query(u, v);
        ans = ans * 2 - a[u] - a[v];
        cout << ans << "\n";
      }
    }
    return 0;
  }
  