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;
  mt19937 rnd(time(0));
  const bool hld_type = 0;
  const int N = 1e5 + 10;
  const int L = 20;
  const long long good_value = 0;
  long long merge(long long a, long long b) { return a + b; }
  struct segment_tree {
    long long t[2 * N + 5];
    void build() {
      for (int i = N - 1; i > 0; i--) {
        t[i] = merge(t[i << 1], t[i << 1 | 1]);
      }
    }
    void update(int pos, long long val) {
      for (t[pos += N] = val; pos > 1; pos >>= 1) {
        t[pos >> 1] = merge(t[pos], t[pos ^ 1]);
      }
    }
    long long get(int l, int r) {
      long long res = good_value;
      for (l += N, r += N; l <= r; l >>= 1, r >>= 1) {
        if (l & 1) res = merge(res, t[l++]);
        if (!(r & 1)) res = merge(res, t[r--]);
      }
      return res;
    }
  } tree;
  int timer, tree_ptr;
  vector<int> g[N];
  int in[N], out[N];
  int up[N][L], down[N];
  int sz[N], depth[N];
  int start[N], st_ptr;
  int path_st[N], pos[N];
  bool is_heavy(int v, int to) { return sz[to] >= (sz[v] + 1) / 2; }
  void dfs_hld(int v = 1, int p = 1) {
    in[v] = ++timer;
    up[v][0] = p;
    for (int i = 1; i < L; i++) {
      up[v][i] = up[up[v][i - 1]][i - 1];
    }
    sz[v] = 1;
    for (int to : g[v]) {
      if (to != p) {
        depth[to] = depth[v] + 1;
        dfs_hld(to, v);
        sz[v] += sz[to];
      }
    }
    bool light = 1;
    for (int to : g[v]) {
      if (to != p && is_heavy(v, to)) {
        light = 0;
        break;
      }
    }
    if (light) {
      start[st_ptr++] = v;
    }
    out[v] = ++timer;
  }
  bool is_anc(int u, int v) { return in[u] <= in[v] && in[v] <= out[u]; }
  int lca(int u, int v) {
    if (is_anc(u, v))
      return u;
    else if (is_anc(v, u))
      return v;
    for (int i = L - 1; i >= 0; i--) {
      if (!is_anc(up[u][i], v)) u = up[u][i];
    }
    return up[u][0];
  }
  int dist(int u, int v) { return depth[u] + depth[v] - 2 * depth[lca(u, v)]; }
  void heavy_light_decomposition(int root = 1) {
    int mem = st_ptr;
    dfs_hld(root, root);
    for (int i = mem; i < st_ptr; i++) {
      int v = start[i];
      while (v != root && is_heavy(up[v][0], v)) {
        down[up[v][0]] = v;
        v = up[v][0];
      }
      int cur_st = v;
      while (v) {
        path_st[v] = cur_st;
        pos[v] = tree_ptr++;
        v = down[v];
      }
    }
  }
  void update(int v, long long val) { tree.update(pos[v], val); }
  long long get(int u, int v, bool flag = 0) {
    if (is_anc(u, v)) {
      long long res = good_value;
      while (path_st[v] != path_st[u]) {
        res = merge(res, tree.get(pos[path_st[v]], pos[v]));
        v = up[path_st[v]][0];
      }
      res = merge(res, tree.get(pos[u] + (hld_type || flag), pos[v]));
      return res;
    } else if (is_anc(v, u)) {
      return get(v, u);
    } else {
      int l = lca(u, v);
      return merge(get(l, u), get(l, v, 1));
    }
  }
  long long a[N];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, q;
    cin >> n >> q;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    heavy_light_decomposition();
    for (int i = 1; i <= n; i++) {
      tree.t[pos[i] + N] = a[i];
    }
    tree.build();
    while (q--) {
      int type;
      cin >> type;
      if (type == 1) {
        int v;
        long long val;
        cin >> v >> val;
        update(v, abs(val));
        a[v] = abs(val);
      } else {
        int u, v;
        cin >> u >> v;
        if (u == v) {
          cout << "0\n";
        } else {
          cout << 2 * get(u, v) - a[u] - a[v] << "\n";
        }
      }
    }
    return 0;
  }
  