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 siz = 1e5 + 40;
  const long long p2 = 1 << 20;
  const long long inf = 1e18;
  vector<long long> graph[siz];
  long long b[siz];
  long long st[siz];
  long long en[siz];
  long long rf[siz];
  long long height[siz];
  struct LCA {
    LCA() {
      t = new pair<long long, long long>[p2 * 2];
      for (long long i = 0; i < p2 * 2; i++)
        t[i] = pair<long long, long long>(inf, inf);
    }
    void upd(long long x, pair<long long, long long> v) {
      x += p2;
      t[x] = v;
      for (x /= 2; x; x /= 2) t[x] = min(t[x * 2], t[x * 2 + 1]);
    }
    long long get(long long b, long long e) {
      b += p2;
      e += p2;
      pair<long long, long long> v = pair<long long, long long>(inf, inf);
      while (b <= e) {
        if (b % 2 == 1) {
          v = min(v, t[b]);
          b++;
        }
        if (e % 2 == 0) {
          v = min(v, t[e]);
          e--;
        }
        b /= 2;
        e /= 2;
      }
      return v.second;
    }
    pair<long long, long long>* t;
  };
  struct Segtree {
    Segtree() {
      t = new long long[p2 * 2];
      for (long long i = 0; i < p2 * 2; i++) t[i] = 0;
    }
    void upd(long long b, long long e, long long v) {
      b += p2;
      e += p2;
      while (b <= e) {
        if (b % 2 == 1) {
          t[b] += v;
          b++;
        }
        if (e % 2 == 0) {
          t[e] += v;
          e--;
        }
        b /= 2;
        e /= 2;
      }
    }
    long long get(long long x) {
      long long sum = 0;
      for (x += p2; x; x /= 2) sum += t[x];
      return sum;
    }
    long long* t;
  };
  long long tim = 0;
  void dfs(long long x, long long p, vector<long long>& path) {
    st[x] = en[x] = tim++;
    path.push_back(x);
    for (long long y : graph[x])
      if (y != p) {
        height[y] = height[x] + 1;
        dfs(y, x, path);
        path.push_back(x);
        en[x] = en[y];
      }
  }
  signed main() {
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    long long n, nbQ;
    cin >> n >> nbQ;
    for (long long i = 0; i < n; i++) {
      cin >> b[i];
      b[i] = max(b[i], -b[i]);
    }
    for (long long i = 0; i < n - 1; i++) {
      long long na, nb;
      cin >> na >> nb;
      na--;
      nb--;
      graph[na].push_back(nb);
      graph[nb].push_back(na);
    }
    vector<long long> path;
    height[0] = 0;
    dfs(0, -1, path);
    LCA lca = LCA();
    for (long long i = 0; i < (long long)(path.size()); i++) {
      lca.upd(i, pair<long long, long long>(height[path[i]], path[i]));
      rf[path[i]] = i;
    }
    Segtree q = Segtree();
    for (long long i = 0; i < n; i++) q.upd(st[i], en[i], b[i]);
    for (long long i = 0; i < nbQ; i++) {
      long long typ;
      cin >> typ;
      if (typ == 1) {
        long long x, v;
        cin >> x >> v;
        v = max(v, -v);
        x--;
        q.upd(st[x], en[x], v - b[x]);
        b[x] = v;
      } else {
        long long x, y;
        cin >> x >> y;
        x--;
        y--;
        long long z = lca.get(min(rf[x], rf[y]), max(rf[x], rf[y]));
        long long loc = q.get(st[x]) + q.get(st[y]) - 2LL * q.get(st[z]);
        loc += b[z];
        loc *= 2LL;
        loc -= b[x];
        loc -= b[y];
        cout << loc;
        cout << '\n';
      }
    }
    cout.flush();
    long long d = 0;
    d++;
  }
  