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;
  template <typename T>
  class BinaryIndexedTree {
   private:
    vector<T> c;
  
   public:
    void Build(int n) { c = vector<T>(n + 1, 0); }
    void Add(int x, T v) {
      for (; x < c.size(); x += x & -x) c[x] += v;
    }
    T Sum(int x) {
      T ret = 0;
      for (; x; x -= x & -x) ret += c[x];
      return ret;
    }
  };
  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);
    }
    int N = 19;
    int dfs_clock = 0;
    vector<int> dep(n + 1), pre(n + 1), sz(n + 1);
    vector<vector<int>> fa(n + 1, vector<int>(N + 1));
    function<void(int)> dfs = [&](int u) {
      for (int i = 1; i <= N; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
      pre[u] = ++dfs_clock;
      sz[u] = 1;
      for (int v : G[u]) {
        if (v == fa[u][0]) continue;
        fa[v][0] = u;
        dep[v] = dep[u] + 1;
        dfs(v);
        sz[u] += sz[v];
      }
    };
    dep[1] = 1;
    dfs(1);
    auto get_lca = [&](int u, int v) {
      if (dep[u] < dep[v]) swap(u, v);
      for (int i = N; i >= 0; i--)
        if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];
      for (int i = N; i >= 0; i--)
        if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
      if (u != v) u = fa[u][0];
      return u;
    };
    BinaryIndexedTree<long long> bit;
    bit.Build(n);
    auto modify = [&](int u, int c) {
      c = abs(c);
      bit.Add(pre[u], c - a[u]);
      bit.Add(pre[u] + sz[u], a[u] - c);
      a[u] = c;
    };
    auto query = [&](int u, int v) {
      if (u == v) return 0LL;
      int lca = get_lca(u, v);
      long long ans = 0;
      ans += bit.Sum(pre[u]) + bit.Sum(pre[v]);
      ans -= bit.Sum(pre[lca]) + bit.Sum(pre[fa[lca][0]]);
      ans *= 2;
      ans -= a[u] + a[v];
      return ans;
    };
    for (int i = 1; i <= n; i++) {
      int w = a[i];
      a[i] = 0;
      modify(i, w);
    }
    for (int i = 0; i < q; i++) {
      int op, x, y;
      cin >> op >> x >> y;
      if (op == 1)
        modify(x, y);
      else
        cout << query(x, y) << '\n';
    }
  }
  int main() {
    cin.tie(0), cout.tie(0);
    ios::sync_with_stdio(false);
    solve();
  }
  