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;
  const int MOD = 1e9 + 7;
  const long long MX = 1e5 + 3;
  const long long INF = 1e9;
  int n;
  int T[100003], L[100003], P[100003][22], s_id[MX], e_id[MX], c;
  vector<int> g[100003];
  void dfs(int from, int u, int dep) {
    s_id[u] = c++;
    T[u] = from;
    L[u] = dep;
    for (auto v : g[u]) {
      if (v == from) continue;
      dfs(u, v, dep + 1);
    }
    e_id[u] = c++;
  }
  void lca_init() {
    c = 1;
    dfs(-1, 1, 0);
    memset(P, -1, sizeof(P));
    for (int i = 1; i <= n; i++) P[i][0] = T[i];
    for (int j = 1; 1 << j <= n; j++) {
      for (int i = 1; i <= n; i++) {
        if (P[i][j - 1] != -1) {
          P[i][j] = P[P[i][j - 1]][j - 1];
        }
      }
    }
  }
  int lca_query(int p, int q) {
    if (L[p] < L[q]) swap(p, q);
    int log = 1;
    while (true) {
      int next = log + 1;
      if (1 << next > L[p]) break;
      log++;
    }
    for (int i = log; i >= 0; i--) {
      if (L[p] - (1 << i) >= L[q]) {
        p = P[p][i];
      }
    }
    if (p == q) return p;
    for (int i = log; i >= 0; i--) {
      if (P[p][i] != -1 && P[p][i] != P[q][i]) {
        p = P[p][i], q = P[q][i];
      }
    }
    return T[p];
  }
  long long bit[MX + MX];
  void update(int i, long long val) {
    for (; i < c; i += (i & (-i))) {
      bit[i] += val;
    }
  }
  long long query(int i) {
    long long ret = 0;
    for (; i > 0; i -= (i & (-i))) {
      ret += bit[i];
    }
    return ret;
  }
  long long sum(int l, int r) { return query(r) - query(l - 1); }
  void solve() {
    int q;
    cin >> n >> q;
    int a[n + 1];
    for (int i = 1; i < n + 1; ++i) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    int u, v;
    for (int i = 1; i < n; ++i) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    lca_init();
    for (int i = 1; i < n + 1; ++i) {
      update(s_id[i], a[i]);
      update(e_id[i], -a[i]);
    }
    int tp, _a;
    while (q--) {
      cin >> tp;
      if (tp == 1) {
        cin >> u >> _a;
        update(s_id[u], -a[u]);
        update(e_id[u], a[u]);
        a[u] = abs(_a);
        update(s_id[u], a[u]);
        update(e_id[u], -a[u]);
      } else {
        cin >> u >> v;
        int lca = lca_query(u, v);
        long long ans;
        if (u == v)
          ans = 0;
        else if (u == lca || v == lca) {
          if (u == lca) swap(u, v);
          ans = 2 * (sum(s_id[1], s_id[u]) - sum(s_id[1], s_id[lca]));
          ans += a[lca];
          ans -= a[u];
        } else {
          ans = sum(s_id[1], s_id[u]) + sum(s_id[1], s_id[v]) -
                (2 * sum(s_id[1], s_id[lca])) + a[lca];
          ans = 2 * ans;
          ans -= a[u];
          ans -= a[v];
        }
        cout << ans << '\n';
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    for (int t1 = 1; t1 < t + 1; ++t1) {
      solve();
    }
    return 0;
  }
  