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 int maxn = 100005;
  vector<int> g[maxn];
  int f[maxn][20], dep[maxn], a[maxn], id[maxn], sz[maxn], dcnt = 0;
  void dfs(int u, int fa) {
    id[u] = ++dcnt;
    sz[u] = 1;
    for (int v : g[u]) {
      if (v == fa) continue;
      dep[v] = dep[u] + 1;
      f[v][0] = u;
      for (int i = 1; i <= 19; ++i) f[v][i] = f[f[v][i - 1]][i - 1];
      dfs(v, u);
      sz[u] += sz[v];
    }
  }
  int lca(int u, int v) {
    if (dep[u] < dep[v]) swap(u, v);
    for (int d = dep[u] - dep[v], i = 19; i >= 0; --i)
      if (d & (1 << i)) u = f[u][i];
    if (u == v) return u;
    for (int i = 19; i >= 0; --i)
      if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];
    return f[u][0];
  }
  long long s[maxn << 2], lz[maxn << 2];
  void update(int L, int R, int v, int l, int r, int rt) {
    if (L <= l && r <= R) {
      s[rt] += 1ll * v * (r - l + 1);
      lz[rt] += v;
      return;
    }
    int m = (l + r) / 2;
    if (lz[rt]) {
      lz[rt << 1] += lz[rt];
      lz[rt << 1 | 1] += lz[rt];
      s[rt << 1] += lz[rt] * (m - l + 1);
      s[rt << 1 | 1] += lz[rt] * (r - m);
      lz[rt] = 0;
    }
    if (L <= m) update(L, R, v, l, m, rt << 1);
    if (m < R) update(L, R, v, m + 1, r, rt << 1 | 1);
    s[rt] = s[rt << 1] + s[rt << 1 | 1];
  }
  long long query(int p, int l, int r, int rt) {
    if (l == r) return s[rt];
    int m = (l + r) / 2;
    if (lz[rt]) {
      lz[rt << 1] += lz[rt];
      lz[rt << 1 | 1] += lz[rt];
      s[rt << 1] += lz[rt] * (m - l + 1);
      s[rt << 1 | 1] += lz[rt] * (r - m);
      lz[rt] = 0;
    }
    if (p <= m) return query(p, l, m, rt << 1);
    return query(p, m + 1, r, rt << 1 | 1);
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, q;
    cin >> n >> q;
    for (int i = 1; i <= n; ++i) cin >> a[i], a[i] = abs(a[i]);
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; ++i) update(id[i], id[i] + sz[i] - 1, a[i], 1, n, 1);
    while (q--) {
      int op, u, v;
      cin >> op >> u >> v;
      if (op == 1) {
        update(id[u], id[u] + sz[u] - 1, -a[u], 1, n, 1);
        a[u] = abs(v);
        update(id[u], id[u] + sz[u] - 1, a[u], 1, n, 1);
      } else {
        int p = lca(u, v);
        long long ans = query(id[u], 1, n, 1) + query(id[v], 1, n, 1);
        if (p != 1) ans -= 2 * query(id[f[p][0]], 1, n, 1);
        ans -= a[p];
        ans *= 2;
        ans -= a[u], ans -= a[v];
        cout << ans << '\n';
      }
    }
  }
  