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;
  using namespace std;
  const long long N = 1e5 + 5;
  const long long D = 19;
  const long long S = (1 << D);
  long long n, q, v[N];
  vector<long long> adj[N];
  long long sz[N], p[N], dep[N];
  long long st[S], id[N], tp[N];
  void update(long long idx, long long val) {
    st[idx += n] = val;
    for (idx /= 2; idx; idx /= 2) st[idx] = st[2 * idx] + st[2 * idx + 1];
  }
  long long query(long long lo, long long hi) {
    long long ra = 0, rb = 0;
    for (lo += n, hi += n + 1; lo < hi; lo /= 2, hi /= 2) {
      if (lo & 1) ra += st[lo++];
      if (hi & 1) rb += st[--hi];
    }
    return ra + rb;
  }
  long long dfs_sz(long long cur, long long par) {
    sz[cur] = 1;
    p[cur] = par;
    for (long long chi : adj[cur]) {
      if (chi == par) continue;
      dep[chi] = dep[cur] + 1;
      p[chi] = cur;
      sz[cur] += dfs_sz(chi, cur);
    }
    return sz[cur];
  }
  long long ct = 1;
  void dfs_hld(long long cur, long long par, long long top) {
    id[cur] = ct++;
    tp[cur] = top;
    update(id[cur], v[cur]);
    long long h_chi = -1, h_sz = -1;
    for (long long chi : adj[cur]) {
      if (chi == par) continue;
      if (sz[chi] > h_sz) {
        h_sz = sz[chi];
        h_chi = chi;
      }
    }
    if (h_chi == -1) return;
    dfs_hld(h_chi, cur, top);
    for (long long chi : adj[cur]) {
      if (chi == par || chi == h_chi) continue;
      dfs_hld(chi, cur, chi);
    }
  }
  long long path(long long x, long long y) {
    long long ret = 0;
    while (tp[x] != tp[y]) {
      if (dep[tp[x]] < dep[tp[y]]) swap(x, y);
      ret += query(id[tp[x]], id[x]);
      x = p[tp[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ret += query(id[x], id[y]);
    return ret;
  }
  signed main() {
    cin.tie(0)->sync_with_stdio(0);
    cin >> n >> q;
    for (long long i = 1; i <= n; i++) {
      cin >> v[i];
      v[i] = abs(v[i]);
    }
    for (long long i = 2; i <= n; i++) {
      long long a, b;
      cin >> a >> b;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    dfs_sz(1, 1);
    dfs_hld(1, 1, 1);
    while (q--) {
      long long t;
      cin >> t;
      if (t == 1) {
        long long s, x;
        cin >> s >> x;
        v[s] = abs(x);
        update(id[s], v[s]);
      } else {
        long long a, b;
        cin >> a >> b;
        long long res = 2 * path(a, b) - v[a] - v[b];
        cout << res << "\n";
      }
    }
  }
  