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;
  template <class T>
  bool umin(T &a, const T &b) {
    return (a > b ? a = b, 1 : 0);
  }
  template <class T>
  bool umax(T &a, const T &b) {
    return (a < b ? a = b, 1 : 0);
  }
  void INPUT(bool F = 0) {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    if (F) {
      freopen(
          "name"
          ".in",
          "r", stdin);
      freopen(
          "name"
          ".out",
          "w", stdout);
    }
  }
  const int mod = (int)1e9 + 7;
  const int N = (int)1e5 + 15;
  const int LG = (int)17;
  vector<int> g[N];
  int a[N], up[N][LG], tin[N], tout[N], T = 2;
  long long h[N], fen[N];
  bool anc(int u, int v) { return (tin[u] <= tin[v]) && (tout[u] >= tout[v]); }
  void dfs(int v, int p, long long H) {
    up[v][0] = p;
    tin[v] = T++;
    H += a[v];
    h[v] = H;
    for (int i = 1; i < LG; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
    for (auto u : g[v]) {
      if (u != p) {
        dfs(u, v, H);
      }
    }
    tout[v] = T - 1;
  }
  void modify(int x, long long pl) {
    for (; x < N; x += x & -x) fen[x] += pl;
  }
  long long get(int r) {
    long long ans = 0;
    for (; r > 0; r -= r & -r) {
      ans += fen[r];
    }
    return ans;
  }
  int get_lca(int u, int v) {
    if (anc(u, v)) return u;
    if (anc(v, u)) return v;
    for (int i = LG - 1; i >= 0; --i) {
      if (!anc(up[u][i], v)) u = up[u][i];
    }
    return up[u][0];
  }
  void change(int v, int val) {
    val = abs(val);
    int diff = val - a[v];
    a[v] = val;
    modify(tin[v], +diff);
    modify(tout[v] + 1, -diff);
  }
  long long gh(int v) { return get(tin[v]); }
  long long query(int u, int v) {
    if (anc(v, u)) swap(v, u);
    if (anc(u, v)) {
      long long ans = (gh(v) - gh(u)) * 2 - a[v] + a[u];
      return ans;
    }
    int lca = get_lca(u, v);
    long long ans =
        (gh(u) - gh(lca)) * 2 + (gh(v) - gh(lca)) * 2 - a[u] - a[v] + 2 * a[lca];
    return ans;
  }
  signed main() {
    INPUT();
    int n, q;
    cin >> n >> q;
    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;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(0, 0, 0);
    for (int i = 0; i < n; ++i) {
      modify(tin[i], h[i]);
      modify(tin[i] + 1, -h[i]);
    }
    for (int i = 0; i < q; ++i) {
      int t, u, c;
      cin >> t >> u >> c;
      --u;
      if (t == 1) {
        change(u, c);
      } else {
        --c;
        cout << query(u, c) << '\n';
      }
    }
    return 0;
  }
  