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 maxn = 1e5 + 5;
  long long n, q;
  long long deep[maxn], fa[maxn], top[maxn], siz[maxn], son[maxn], id[maxn];
  vector<long long> ve[maxn];
  long long cnt = 0;
  long long a[maxn];
  long long dfs1(long long u, long long f, long long d) {
    deep[u] = d;
    fa[u] = f;
    siz[u] = 1;
    for (auto v : ve[u]) {
      if (v == f) continue;
      dfs1(v, u, d + 1);
      siz[u] += siz[v];
      if (siz[v] > siz[son[u]]) son[u] = v;
    }
    return 0;
  }
  long long dfs2(long long u, long long topf) {
    id[u] = ++cnt;
    top[u] = topf;
    if (!son[u]) return 0;
    dfs2(son[u], topf);
    for (auto v : ve[u]) {
      if (v == fa[u] || v == son[u]) continue;
      dfs2(v, v);
    }
    return 0;
  }
  const long long maxm = maxn << 2;
  long long szsz[maxm];
  long long update(long long x, long long y) {
    while (x < maxm) {
      szsz[x] += y;
      x += (x & (-x));
    }
    return 0;
  }
  long long query(long long x) {
    long long ret = 0;
    while (x) {
      ret += szsz[x];
      x -= (x & (-x));
    }
    return ret;
  }
  long long val[maxn];
  inline long long calc(long long x, long long y) {
    return max(abs(x + y), abs(x - y));
  }
  long long dfs3(long long u, long long f) {
    for (auto v : ve[u]) {
      if (v == f) continue;
      val[v] = calc(a[u], a[v]);
      dfs3(v, u);
    }
    return 0;
  }
  long long qrange(long long x, long long y) {
    if (x == y) return 0;
    long long ans = 0;
    while (top[x] != top[y]) {
      if (deep[top[x]] < deep[top[y]]) swap(x, y);
      if (x != top[x]) {
        ans += query(id[x]) - query(id[top[x]]);
        ans += calc(a[top[x]], a[fa[top[x]]]);
      } else {
        ans += calc(a[x], a[fa[x]]);
      }
      x = fa[top[x]];
    }
    if (deep[x] > deep[y]) swap(x, y);
    ans += query(id[y]) - query(id[x]);
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> q;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 1; i <= n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      ve[u].push_back(v);
      ve[v].push_back(u);
    }
    dfs1(1, -1, 1);
    dfs2(1, 1);
    dfs3(1, -1);
    for (long long i = 1; i <= n; i++) {
      update(id[i], val[i]);
    }
    for (long long i = 1; i <= q; i++) {
      long long op, s, t;
      cin >> op >> s >> t;
      if (op == 1) {
        a[s] = t;
        if (son[s]) {
          long long val = query(id[son[s]]) - query(id[s]);
          update(id[son[s]], -val);
          update(id[son[s]], calc(a[s], a[son[s]]));
        }
        if (fa[s] != -1 && son[fa[s]] == s) {
          long long val = query(id[s]) - query(id[fa[s]]);
          update(id[s], -val);
          update(id[s], calc(a[s], a[fa[s]]));
        }
      } else {
        cout << qrange(s, t) << '\n';
      }
    }
    return 0;
  }
  