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 N, Left, Right, Value;
  long long seg[2000001], lazy[2000001], sparse[200001][20], First[200001],
      Last[200001], A[200001], depth[200001], dist[200001];
  vector<long long> adj[200001];
  void Spread(long long l, long long r, long long ind) {
    if (l != r) {
      lazy[ind * 2] += lazy[ind];
      seg[ind * 2] += lazy[ind];
      lazy[ind * 2 + 1] += lazy[ind];
      seg[ind * 2 + 1] += lazy[ind];
    }
    lazy[ind] = 0;
  }
  long long Query(long long l = 1, long long r = N, long long ind = 1) {
    if (l > Right || r < Left) return 0;
    Spread(l, r, ind);
    if (l >= Left and r <= Right) return seg[ind];
    long long mid = (l + r) / 2;
    return Query(l, mid, ind * 2) + Query(mid + 1, r, ind * 2 + 1);
  }
  void Update(long long l = 1, long long r = N, long long ind = 1) {
    if (l > Right || r < Left) return;
    Spread(l, r, ind);
    if (l >= Left and r <= Right) {
      lazy[ind] += Value;
      seg[ind] += Value;
      return;
    }
    long long mid = (l + r) / 2;
    Update(l, mid, ind * 2);
    Update(mid + 1, r, ind * 2 + 1);
    seg[ind] = seg[ind * 2] + seg[ind * 2 + 1];
  }
  long long pos(long long x) {
    Left = Right = First[x];
    return Query();
  }
  long long ind = 1;
  void dfs(long long cur = 1, long long par = 0, long long dep = 0) {
    sparse[cur][0] = par;
    depth[cur] = dep;
    First[cur] = ind++;
    if (cur != 1) dist[cur] = dist[par] + A[cur] + A[par];
    for (long long i = 1; (1 << i) <= dep; i++) {
      sparse[cur][i] = sparse[sparse[cur][i - 1]][i - 1];
    }
    for (long long i : adj[cur]) {
      if (i != par) {
        dfs(i, cur, dep + 1);
      }
    }
    Last[cur] = ind - 1;
  }
  long long kth(long long a, long long k) {
    for (long long i = 19; i >= 0; i--) {
      if (k & (1 << i)) a = sparse[a][i];
    }
    return a;
  }
  long long lca(long long a, long long b) {
    if (depth[b] > depth[a]) swap(a, b);
    long long d = depth[a] - depth[b];
    a = kth(a, d);
    if (a == b) return a;
    for (long long i = 19; i >= 0; i--) {
      if (sparse[a][i] != sparse[b][i]) {
        a = sparse[a][i];
        b = sparse[b][i];
      }
    }
    return sparse[a][0];
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long n, q;
    cin >> n >> q;
    N = exp2(ceil(log2(n)));
    for (long long i = 1; i <= n; i++) {
      cin >> A[i];
      A[i] = abs(A[i]);
    }
    for (long long i = 1; i < n; i++) {
      long long a, b;
      cin >> a >> b;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    dfs();
    for (long long i = 1; i <= n; i++) {
      seg[N + First[i] - 1] = dist[i];
    }
    while (q--) {
      long long t, a, b;
      cin >> t >> a >> b;
      if (--t) {
        cout << pos(a) + pos(b) - 2 * pos(lca(a, b)) << '\n';
      } else {
        b = abs(b);
        if (a != 1) {
          Left = Right = First[a];
          Value = -A[a] + b;
          Update();
          Left = First[a] + 1;
          Right = Last[a];
          Value = 2 * (-A[a] + b);
          Update();
        } else {
          Left = First[a] + 1;
          Right = Last[a];
          Value = (-A[a] + b);
          Update();
        }
        A[a] = b;
      }
    }
  }
  