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 + 7;
  const long long inf = 0x3f3f3f3f;
  const int LOG = 20;
  int n, q, timer, w[N], in[N], out[N];
  long long a[N], T[N];
  vector<int> adj[N];
  void update(int pos, int v) {
    for (; pos <= n; pos += pos & -pos) T[pos] += v;
  }
  long long get(int pos) {
    long long res = 0;
    for (; pos > 0; pos -= pos & -pos) res += T[pos];
    return res;
  }
  int h[N], P[N][LOG];
  void dfs(int u, int p) {
    in[u] = ++timer;
    for (int v : adj[u])
      if (v ^ p) h[v] = h[u] + 1, P[v][0] = u, dfs(v, u);
    out[u] = timer;
  }
  void buildLCA(int s) {
    h[s] = 0, dfs(s, s);
    for (int j = 1; (1 << j) <= n; j++)
      for (long long i = (1); i < (n + 1); i++)
        if (P[i][j - 1]) P[i][j] = P[P[i][j - 1]][j - 1];
  }
  int LCA(int u, int v) {
    if (h[u] < h[v]) swap(u, v);
    int hu = h[u], hv = h[v];
    int log = log2(h[u]);
    for (long long i = (log); i >= (0); i--)
      if (h[u] - (1 << i) >= h[v]) u = P[u][i];
    if (u == v) return u;
    for (long long i = (log); i >= (0); i--)
      if (P[u][i] && P[u][i] != P[v][i]) u = P[u][i], v = P[v][i];
    return P[u][0];
  }
  int main() {
    ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    cin >> n >> q;
    for (long long i = (1); i < (n + 1); i++) cin >> a[i], a[i] = abs(a[i]);
    for (long long i = (1); i < (n); i++) {
      int u, v;
      cin >> u >> v;
      adj[u].push_back(v), adj[v].push_back(u);
    }
    buildLCA(1);
    for (long long i = (1); i < (n + 1); i++)
      update(in[i], a[i]), update(out[i] + 1, -a[i]);
    while (q--) {
      int type, u, v;
      cin >> type >> u >> v;
      if (type == 1) {
        update(in[u], -a[u]), update(out[u] + 1, a[u]);
        a[u] = abs(v);
        update(in[u], a[u]), update(out[u] + 1, -a[u]);
      } else {
        int s = LCA(u, v);
        long long ans = 2 * (get(in[u]) + get(in[v]) - 2 * get(in[s])) +
                        2 * a[s] - a[u] - a[v];
        cout << ans << endl;
      }
    }
  }
  