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;
  struct SegTree {
   private:
    long long n;
    vector<long long> t;
    long long query(long long id, long long l, long long r, long long l1,
                    long long r1) {
      if (l1 == r1) return 0;
      if (r == r1 and l == l1) return t[id];
      long long m = (l + r) >> 1;
      return query(id << 1, l, m, min(l1, m), min(r1, m)) +
             query((id << 1) + 1, m, r, max(l1, m), max(r1, m));
    }
    long long update(long long id, long long l, long long r, long long i,
                     long long del) {
      if (i < l or i >= r) return t[id];
      if (r - l == 1) return t[id] += del;
      long long m = (l + r) >> 1;
      return t[id] = update(id << 1, l, m, i, del) +
                     update((id << 1) + 1, m, r, i, del);
    }
  
   public:
    long long query(long long l, long long r) { return query(1, 0, n, l, r); }
    void update(long long i, long long del) { update(1, 0, n, i, del); }
    SegTree(long long n) {
      this->n = n;
      t.assign(4 * n, 0);
    }
  };
  long long n, l;
  vector<vector<long long>> adj;
  vector<long long> a;
  long long timer;
  vector<long long> tin, tout;
  vector<vector<long long>> up;
  void dfs(long long v, long long p) {
    tin[v] = timer++;
    up[v][0] = p;
    for (long long i = 1; i <= l; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
    for (long long u : adj[v]) {
      if (u != p) dfs(u, v);
    }
    tout[v] = timer++;
  }
  bool is_ancestor(long long u, long long v) {
    return tin[u] <= tin[v] && tout[u] >= tout[v];
  }
  long long lca(long long u, long long v) {
    if (is_ancestor(u, v)) return u;
    if (is_ancestor(v, u)) return v;
    for (long long i = l; i >= 0; --i) {
      if (!is_ancestor(up[u][i], v)) u = up[u][i];
    }
    return up[u][0];
  }
  void preprocess(long long root) {
    tin.resize(n);
    tout.resize(n);
    timer = 0;
    l = ceil(log2(n));
    up.assign(n, vector<long long>(l + 1));
    dfs(root, root);
  }
  void update(long long u, long long k, SegTree& p) {
    k = k < 0 ? -k : k;
    p.update(tin[u], k - a[u]);
    p.update(tout[u], a[u] - k);
    a[u] = k;
  }
  long long query(long long u, long long v, SegTree& p) {
    long long anc = lca(u, v);
    long long x = p.query(0, tout[u]);
    long long y = p.query(0, tout[v]);
    long long z = p.query(0, tout[anc]);
    return 2 * (x + y - 2 * z + a[anc]) - a[u] - a[v];
  }
  signed main() {
    iostream::sync_with_stdio(false);
    long long q;
    cin >> n >> q;
    vector<long long> a1(n);
    for (long long& a11 : a1) cin >> a11;
    adj.assign(n, vector<long long>());
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      u--;
      v--;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    preprocess(0);
    SegTree p(timer);
    a.assign(n, 0);
    for (long long u = 0; u < n; u++) update(u, a1[u], p);
    while (q--) {
      long long x, y, z;
      cin >> x >> y >> z;
      if (x == 1)
        update(y - 1, z, p);
      else
        cout << query(y - 1, z - 1, p) << '\n';
      ;
    }
    assert(cin);
  }
  