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 long long Mod = 1e9 + 7, N = 200005;
  long long tree[N << 1];
  void add(int idx, long long val) {
    while (idx <= 2 * N) {
      tree[idx] += val;
      idx += idx & (-idx);
    }
  }
  long long get(int idx) {
    long long ret = 0;
    while (idx > 0) {
      ret += tree[idx];
      idx -= idx & (-idx);
    }
    return ret;
  }
  int up[22][N], depth[N];
  vector<int> v[N];
  int sz = 1;
  int l[N], f[N];
  long long c[N];
  void dfs(int u, int p) {
    f[u] = sz;
    sz++;
    up[0][u] = p;
    for (auto r : v[u]) {
      if (r != p) {
        depth[r] = depth[u] + 1;
        dfs(r, u);
      }
    }
    l[u] = sz;
    sz++;
  }
  int get_LCA(int a, int b) {
    if (depth[a] < depth[b]) swap(a, b);
    int k = depth[a] - depth[b];
    for (int j = 18; j >= 0; j--) {
      if ((k >> j) & 1) a = up[j][a];
    }
    if (a == b) return a;
    for (int j = 18; j >= 0; j--) {
      if (up[j][a] != up[j][b]) {
        a = up[j][a];
        b = up[j][b];
      }
    }
    return up[0][a];
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t, q, x, n, y;
    t = 1;
    for (int w = 1; w <= t; w++) {
      cin >> n >> q;
      for (int i = 1; i <= n; i++) {
        cin >> c[i];
        c[i] = abs(c[i]);
      }
      for (int i = 0; i < n - 1; i++) {
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
      }
      sz = 1;
      depth[1] = 1;
      dfs(1, 1);
      for (int j = 1; j <= 18; j++) {
        for (int i = 1; i <= n; i++) {
          up[j][i] = up[j - 1][up[j - 1][i]];
        }
      }
      cout << '\n';
      for (int i = 1; i <= n; i++) {
        add(f[i], c[i]);
        add(l[i], -c[i]);
      }
      while (q--) {
        cin >> x;
        if (x == 2) {
          cin >> x >> y;
          int z = get_LCA(x, y);
          long long ans = get(f[x]) + get(f[y]) - 2 * get(f[z]) + c[z];
          if (z != x && z != y) {
            x = up[0][x];
            y = up[0][y];
            int zz = get_LCA(x, y);
            ans += get(f[x]) + get(f[y]) - 2 * get(f[z]) + c[z];
            cout << ans << '\n';
          } else if (x == z) {
            ans += get(f[up[0][y]]) - get(f[x]);
            cout << ans << '\n';
          } else if (y == z) {
            ans += get(f[up[0][x]]) - get(f[y]);
            cout << ans << '\n';
          }
        } else {
          cin >> x >> y;
          y = abs(y);
          int pre = c[x];
          c[x] = y;
          add(f[x], y - pre);
          add(l[x], pre - y);
        }
      }
    }
    return 0;
  }
  