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;
  vector<int> edges[100005];
  int parent[100005][20], sz[100005], depth[100005], in[100005], out[100005], sn;
  bool visited[100005];
  int t = 0;
  void init(int s) {
    memset(visited, 0, sizeof(visited));
    sn = s;
  }
  int dfs(int u = 0) {
    sz[u] = 1;
    visited[u] = true;
    in[u] = t++;
    for (int i = 0; i < edges[u].size(); i++) {
      int c = edges[u][i];
      if (visited[c]) continue;
      visited[c] = true;
      parent[c][0] = u;
      depth[c] = depth[u] + 1;
      sz[u] += dfs(c);
    }
    out[u] = t;
    return sz[u];
  }
  void preCompSparseMatrix() {
    for (int i = 1; i < 20; i++) {
      for (int j = 0; j < sn; j++) {
        parent[j][i] = parent[parent[j][i - 1]][i - 1];
      }
    }
  }
  int LCA(int u, int v) {
    if (depth[v] < depth[u]) return LCA(v, u);
    int dif = depth[v] - depth[u];
    for (int i = 0; i < 20; i++) {
      if ((dif >> i) & 1) {
        v = parent[v][i];
      }
    }
    if (u == v) {
      return u;
    }
    for (int i = 19; i >= 0; i--) {
      if (parent[u][i] != parent[v][i]) {
        u = parent[u][i];
        v = parent[v][i];
      }
    }
    return parent[u][0];
  }
  long long st[262144];
  void update(int x, long long v, int l, int r, int p) {
    if (l > x || r < x) return;
    st[p] += v;
    if (l == r) return;
    int m = (l + r) / 2;
    update(x, v, l, m, 2 * p + 1);
    update(x, v, m + 1, r, 2 * p + 2);
    st[p] = st[2 * p + 1] + st[2 * p + 2];
    return;
  }
  void update(int x, long long v) { update(x, v, 0, 100005, 0); }
  long long getSum(int a, int b, int l, int r, int p) {
    if (l > b || r < a) return 0;
    if (l >= a && r <= b) return st[p];
    int m = (l + r) / 2;
    return getSum(a, b, l, m, 2 * p + 1) + getSum(a, b, m + 1, r, 2 * p + 2);
  }
  long long getSum(int a, int b) { return getSum(a, b, 0, 100005, 0); }
  int N, Q;
  long long arr[100005];
  long long get(int a, int b) {
    int l = LCA(a, b);
    return getSum(0, in[a]) + getSum(0, in[b]) - 2ll * getSum(0, in[l]) + arr[l];
  }
  void change(int a, long long b) {
    update(in[a], b);
    update(out[a], -b);
    return;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> N >> Q;
    for (int i = 0; i < N; ++i) {
      cin >> arr[i];
      arr[i] = abs(arr[i]);
    }
    for (int i = 0; i < N - 1; ++i) {
      int a, b;
      cin >> a >> b;
      a--;
      b--;
      edges[a].push_back(b);
      edges[b].push_back(a);
    }
    init(N);
    dfs(0);
    preCompSparseMatrix();
    for (int i = 0; i < N; ++i) {
      change(i, arr[i]);
    }
    for (int i = 0; i < Q; ++i) {
      int a, b, c;
      cin >> a >> b >> c;
      c = abs(c);
      if (a == 1) {
        b--;
        change(b, c - arr[b]);
        arr[b] = c;
      } else {
        b--;
        c--;
        cout << 2ll * get(b, c) - arr[b] - arr[c] << "\n";
      }
    }
    return 0;
  }
  