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 N = 18;
  struct zkw_segmentree {
    int M;
    long long sumv[2 << N];
    void Init(int n) {
      for (M = 1; M < n; M <<= 1)
        ;
      for (int i = 1; i < M << 1; i++) sumv[i] = 0;
    }
    void Update(int u, long long v) {
      for (u += M; u; u >>= 1) sumv[u] += v;
    }
    long long Query(int u, int v) {
      long long ret = 0;
      if (v < u) return ret;
      for (u += M - 1, v += M + 1; u ^ v ^ 1; u >>= 1, v >>= 1) {
        if (~u & 1) ret += sumv[u ^ 1];
        if (v & 1) ret += sumv[v ^ 1];
      }
      return ret;
    }
  } tree;
  void solve() {
    int n, q;
    cin >> n >> q;
    vector<long long> a(n + 1);
    vector<vector<int>> G(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      G[x].push_back(y);
      G[y].push_back(x);
    }
    vector<int> son(n + 1), dep(n + 1), max_dep(n + 1), fa(n + 1);
    function<void(int)> dfs1 = [&](int u) {
      max_dep[u] = dep[u];
      for (int v : G[u]) {
        if (v == fa[u]) continue;
        fa[v] = u;
        dep[v] = dep[u] + 1;
        dfs1(v);
        if (max_dep[v] > max_dep[u]) {
          son[u] = v;
          max_dep[u] = max_dep[v];
        }
      }
    };
    dep[1] = 0;
    dfs1(1);
    int dfs_clock = 0;
    vector<int> pre(n + 1), top(n + 1);
    function<void(int u)> dfs2 = [&](int u) {
      pre[u] = ++dfs_clock;
      if (son[u]) {
        top[son[u]] = top[u];
        dfs2(son[u]);
      }
      for (int v : G[u]) {
        if (v == fa[u] || v == son[u]) continue;
        top[v] = v;
        dfs2(v);
      }
    };
    top[1] = 1;
    dfs2(1);
    auto cost_f = [&](int x, int y) { return max(abs(x + y), abs(x - y)); };
    tree.Init(n);
    for (int i = 1; i <= n; i++) {
      if (son[i]) {
        tree.Update(pre[i], cost_f(a[i], a[son[i]]));
      }
    }
    for (int i = 0; i < q; i++) {
      int op, x, y;
      cin >> op >> x >> y;
      if (op == 1) {
        if (son[x]) {
          long long dt = cost_f(y, a[son[x]]) - cost_f(a[x], a[son[x]]);
          tree.Update(pre[x], dt);
        }
        if (top[x] != x) {
          long long dt = cost_f(y, a[fa[x]]) - cost_f(a[x], a[fa[x]]);
          tree.Update(pre[fa[x]], dt);
        }
        a[x] = y;
      } else {
        long long ans = 0;
        while (top[x] != top[y]) {
          if (dep[top[x]] < dep[top[y]]) swap(x, y);
          ans += tree.Query(pre[top[x]], pre[x] - 1);
          ans += cost_f(a[fa[top[x]]], a[top[x]]);
          x = fa[top[x]];
        }
        if (dep[x] > dep[y]) swap(x, y);
        ans += tree.Query(pre[x], pre[y] - 1);
        cout << ans << '\n';
      }
    }
  }
  int main() {
    cin.tie(0), cout.tie(0);
    ios::sync_with_stdio(false);
    solve();
  }
  