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;
  int n, q;
  int a[100005];
  int de[100005];
  int pr[100005][20];
  int sa[100005], so[100005];
  int cl[100005];
  long long b[100005];
  int id = 0;
  vector<int> g[100005];
  void update(int u, int delta) {
    while (u <= n) b[u] += delta, u += u & (-u);
  }
  long long get(int u) {
    long long res = 0;
    while (u > 0) {
      res += b[u];
      u -= u & (-u);
    }
    return res;
  }
  void DFS(int u) {
    cl[u] = 1;
    sa[u] = ++id;
    for (auto &v : g[u])
      if (!cl[v]) pr[v][0] = u, de[v] = de[u] + 1, DFS(v);
    so[u] = id;
  }
  int LCA(int u, int v) {
    while (de[u] > de[v]) {
      int h = log2(de[u] - de[v]);
      u = pr[u][h];
    }
    while (de[v] > de[u]) {
      int h = log2(de[v] - de[u]);
      v = pr[v][h];
    }
    if (u == v) return u;
    int h = log2(de[u]);
    while (h >= 0) {
      if (pr[u][h] != pr[v][h]) u = pr[u][h], v = pr[v][h];
      h--;
    }
    u = pr[u][0];
    return u;
  }
  int main() {
    ios ::sync_with_stdio(false);
    cin >> n >> q;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    de[1] = 1;
    DFS(1);
    int k = log2(n);
    for (int i = 1; i <= k; ++i)
      for (int j = 1; j <= n; ++j) pr[j][i] = pr[pr[j][i - 1]][i - 1];
    for (int i = 1; i <= n; ++i) update(sa[i], a[i]), update(so[i] + 1, -a[i]);
    for (int i = 1; i <= q; ++i) {
      int tp, u, v;
      cin >> tp >> u >> v;
      if (tp == 1) {
        update(sa[u], -a[u]), update(so[u] + 1, a[u]);
        a[u] = abs(v);
        update(sa[u], a[u]), update(so[u] + 1, -a[u]);
      } else {
        int z = LCA(u, v);
        long long ans = get(sa[u]) + get(sa[v]);
        ans -= 2 * get(sa[z]);
        ans *= 2;
        ans += 2 * a[z];
        ans -= a[u] + a[v];
        cout << ans << "\n";
      }
    }
  }
  