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;
  long long in_time[200005];
  long long a[200005];
  long long out_time[200005];
  long long depth[200005];
  long long tim = 1, n;
  long long bit[2000005];
  vector<long long> adj[200005];
  long long par[200005][25];
  void upd(long long ind, long long v, long long n) {
    while (ind <= n) {
      bit[ind] += v;
      ind += (ind & (-ind));
    }
  }
  long long query(long long ind) {
    long long s = 0;
    while (ind > 0) {
      s += bit[ind];
      ind -= (ind & (-ind));
    }
    return s;
  }
  void dfs(long long v, long long p) {
    depth[v] = depth[p] + 1;
    par[v][0] = p;
    in_time[v] = tim++;
    upd(in_time[v], a[v], 2 * n);
    for (long long i = 1; i <= 20; i++) {
      par[v][i] = par[par[v][i - 1]][i - 1];
    }
    for (auto to : adj[v]) {
      if (to == p) continue;
      dfs(to, v);
    }
    out_time[v] = tim++;
    upd(out_time[v], -a[v], 2 * n);
  }
  long long lca(long long u, long long v) {
    if (depth[u] > depth[v]) {
      swap(u, v);
    }
    long long diff = abs(depth[u] - depth[v]);
    for (long long i = 0; i <= 20; i++) {
      if (diff & (1ll << i)) {
        v = par[v][i];
      }
    }
    if (u == v) {
      return v;
    }
    for (long long i = 20; i >= 0; i--) {
      if (par[u][i] == par[v][i]) {
        continue;
      }
      u = par[u][i];
      v = par[v][i];
    }
    return par[u][0];
  }
  void solve(long long tc) {
    long long q, x, y;
    cin >> n >> q;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (long long i = 1; i < n; i++) {
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    dfs(1, 0);
    long long ty;
    while (q--) {
      cin >> ty >> x >> y;
      if (ty == 1) {
        upd(in_time[x], -a[x], 2 * n);
        upd(out_time[x], a[x], 2 * n);
        a[x] = abs(y);
        upd(in_time[x], a[x], 2 * n);
        upd(out_time[x], -a[x], 2 * n);
      } else {
        long long lc = lca(x, y);
        long long d = 2 * (query(in_time[x]) + query(in_time[y]) -
                           2 * query(in_time[lc]) + a[lc]);
        cout << d - a[x] - a[y] << endl;
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    long long t = 1;
    long long tc = 1;
    while (t--) {
      solve(tc);
      tc++;
    }
    return 0;
  }
  