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> pwr(20, 0);
  void dfs(vector<vector<int>> &adj, int p, int c, vector<int> &lca,
           vector<int> &dpth) {
    lca.push_back(c);
    for (auto u : adj[c])
      if (u != p) {
        dpth[u] = dpth[c] + 1;
        dfs(adj, c, u, lca, dpth);
        lca.push_back(c);
      }
  }
  long long func(long long a, long long b) { return a + b; }
  long long sum(int a, int b, int n, vector<long long> &tree) {
    a += n;
    b += n;
    long long s = 0;
    while (a <= b) {
      if (a % 2 == 1) s = func(s, tree[a++]);
      if (b % 2 == 0) s = func(s, tree[b--]);
      a /= 2;
      b /= 2;
    }
    return s;
  }
  void add(int k, long long x, int n, vector<long long> &tree) {
    k += n;
    tree[k] += x;
    for (k /= 2; k >= 1; k /= 2) {
      tree[k] = func(tree[2 * k], tree[2 * k + 1]);
    }
  }
  void create_par(vector<vector<int>> &adj, int p, int c,
                  vector<vector<int>> &par) {
    for (auto ele : adj[c]) {
      if (ele == p) {
        continue;
      }
      par[ele][0] = c;
      for (int i = 1; i < 20; i++) {
        par[ele][i] = par[par[ele][i - 1]][i - 1];
      }
      create_par(adj, c, ele, par);
    }
  }
  int llca(int u, int v, vector<vector<int>> &par, vector<int> &dpth) {
    int j = 19;
    if (dpth[u] > dpth[v]) {
      swap(u, v);
    }
    int dif = dpth[v] - dpth[u];
    for (int i = 0; i < 20; i++) {
      if (pwr[i] & dif) {
        v = par[v][i];
      }
    }
    while (u != v) {
      if (par[u][0] == par[v][0]) {
        u = v = par[u][0];
        continue;
      }
      while (par[u][j] == par[v][j]) {
        j--;
      }
      u = par[u][j], v = par[v][j];
    }
    return u;
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    for (int i = 0; i < 20; i++) {
      pwr[i] = (1 << i);
    }
    int n, q;
    cin >> n >> q;
    vector<vector<int>> adj;
    for (int i = 0; i < n; i++) {
      adj.push_back({});
    }
    vector<long long> a(n, 0);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] < 0) {
        a[i] *= -1;
      }
    }
    for (int i = 0; i < n - 1; i++) {
      int s, t;
      cin >> s >> t;
      s--;
      t--;
      adj[s].push_back(t);
      adj[t].push_back(s);
    }
    vector<int> lca;
    vector<vector<int>> par;
    vector<int> dpth(n, 0);
    for (int i = 0; i < n; i++) {
      par.push_back({});
      for (int j = 0; j < 20; j++) {
        par[i].push_back(0);
      }
    }
    dfs(adj, -1, 0, lca, dpth);
    create_par(adj, -1, 0, par);
    int l = (int)lca.size();
    vector<int> mnpos(n, 1e9), mxpos(n, -1);
    vector<long long> tree(2 * l + 1, 0);
    for (int j = 0; j < l; j++) {
      mnpos[lca[j]] = min(mnpos[lca[j]], j);
      mxpos[lca[j]] = max(mxpos[lca[j]], j);
    }
    for (int i = 0; i < n; i++) {
      add(mnpos[i], a[i], l, tree);
      add(mxpos[i], -a[i], l, tree);
    }
    auto change = [&](int u, long long c) -> void {
      if (c < 0) {
        c *= -1;
      }
      long long prev = a[u];
      a[u] = c;
      add(mnpos[u], a[u] - prev, l, tree);
      add(mxpos[u], -a[u] + prev, l, tree);
    };
    for (int i = 0; i < q; i++) {
      int t;
      cin >> t;
      if (t == 1) {
        int u;
        long long c;
        cin >> u >> c;
        u--;
        change(u, c);
      } else {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        if (u == v) {
          cout << 0 << '\n';
        } else {
          int x = llca(u, v, par, dpth);
          long long ans = -2 * sum(0, mnpos[x], l, tree) +
                          sum(0, mnpos[u], l, tree) + sum(0, mnpos[v], l, tree) +
                          a[x];
          if (mnpos[v] == mxpos[v]) {
            ans += a[v];
          }
          if (mnpos[u] == mxpos[u]) {
            ans += a[u];
          }
          ans *= 2;
          ans -= a[u];
          ans -= a[v];
          cout << ans << '\n';
        }
      }
    }
    return 0;
  }
  