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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  long long bp(long long a, long long b) {
    a %= 1000000007;
    long long res = 1;
    while (b > 0) {
      if (b % 2) res = (res * a) % 1000000007;
      a = (a * a) % 1000000007;
      b /= 2;
    }
    return res;
  }
  const long long inf = 1e9;
  struct Node {
    Node *l = 0, *r = 0;
    long long lo, hi, mset = inf, madd = 0, val = -inf;
    Node(long long lo, long long hi) : lo(lo), hi(hi) {}
    Node(vector<long long>& v, long long lo, long long hi) : lo(lo), hi(hi) {
      if (lo + 1 < hi) {
        long long mid = lo + (hi - lo) / 2;
        l = new Node(v, lo, mid);
        r = new Node(v, mid, hi);
        val = max(l->val, r->val);
      } else
        val = v[lo];
    }
    long long query(long long L, long long R) {
      if (R <= lo || hi <= L) return -inf;
      if (L <= lo && hi <= R) return val;
      push();
      return max(l->query(L, R), r->query(L, R));
    }
    void set(long long L, long long R, long long x) {
      if (R <= lo || hi <= L) return;
      if (L <= lo && hi <= R)
        mset = val = x, madd = 0;
      else {
        push(), l->set(L, R, x), r->set(L, R, x);
        val = max(l->val, r->val);
      }
    }
    void add(long long L, long long R, long long x) {
      if (R <= lo || hi <= L) return;
      if (L <= lo && hi <= R) {
        if (mset != inf)
          mset += x;
        else
          madd += x;
        val += x;
      } else {
        push(), l->add(L, R, x), r->add(L, R, x);
        val = max(l->val, r->val);
      }
    }
    void push() {
      if (!l) {
        long long mid = lo + (hi - lo) / 2;
        l = new Node(lo, mid);
        r = new Node(mid, hi);
      }
      if (mset != inf)
        l->set(lo, hi, mset), r->set(lo, hi, mset), mset = inf;
      else if (madd)
        l->add(lo, hi, madd), r->add(lo, hi, madd), madd = 0;
    }
  };
  long long n, l;
  vector<vector<long long>> adj;
  long long timer;
  vector<long long> tin, tout;
  vector<vector<long long>> up;
  void dfs(long long v, long long p) {
    tin[v] = timer;
    ++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);
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long q;
    cin >> n >> q;
    long long a[n];
    Node tre(0, n);
    for (long long x = 0; x < n; x++) {
      cin >> a[x];
    }
    long long u, v;
    adj.resize(n);
    for (long long x = 0; x < n - 1; x++) {
      cin >> u >> v;
      --u;
      --v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    preprocess(0);
    for (long long x = 0; x < n; x++) {
      tre.add(tin[x], tout[x], abs(a[x]));
    }
    long long c;
    for (long long x = 0; x < q; x++) {
      cin >> c >> u >> v;
      if (c == 1) {
        --u;
        tre.add(tin[u], tout[u], abs(v) - abs(a[u]));
        a[u] = v;
      } else {
        --u;
        --v;
        long long f = lca(u, v);
        long long i = tre.query(tin[u], tin[u] + 1);
        long long j = tre.query(tin[v], tin[v] + 1);
        long long k = tre.query(tin[f], tin[f] + 1);
        long long g = i + j - (2 * k);
        g += abs(a[f]);
        g *= 2;
        g -= (abs(a[u]) + abs(a[v]));
        cout << g << "\n";
      }
    }
  }
  