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>
  const int N = 100005;
  const int M = 200010;
  using namespace std;
  int n, T;
  int fa[N][20], dep[N];
  int cnt = 0, dfn[N], las[N];
  long long a[N], sum[N], val[N];
  int edge[M], lst[N], nxt[M], t = 0;
  int lowbit(int x) { return x & (-x); }
  void UPD(int x, long long num) {
    if (num == 0) return;
    for (; x <= n; x += lowbit(x)) val[x] += num;
  }
  long long GET(int x) {
    long long ans = 0;
    while (x > 0) {
      ans += val[x];
      x -= lowbit(x);
    }
    return ans;
  }
  void ADD(int x, int y) {
    edge[++t] = y;
    nxt[t] = lst[x];
    lst[x] = t;
  }
  void READ() {
    int u, v;
    scanf("%d%d", &n, &T);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      if (a[i] < 0) a[i] = -a[i];
    }
    for (int i = 1; i <= n - 1; i++) {
      scanf("%d%d", &u, &v);
      ADD(u, v);
      ADD(v, u);
    }
  }
  void SEARCH(int x, int pa) {
    dfn[x] = ++cnt;
    fa[x][0] = pa;
    dep[x] = dep[pa] + 1;
    sum[x] = sum[pa] + a[x];
    for (int r = lst[x]; r; r = nxt[r]) {
      if (edge[r] == pa) continue;
      SEARCH(edge[r], x);
    }
    las[x] = cnt;
  }
  void INIT() {
    cnt = 0;
    dep[0] = 0;
    sum[0] = 0;
    SEARCH(1, 0);
    for (int i = 1; i <= 18; i++) {
      for (int x = 1; x <= n; x++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
    }
  }
  long long VAL(int x) { return sum[x] + GET(dfn[x]); }
  int LCA(int x, int y) {
    if (dep[x] > dep[y]) swap(x, y);
    int de = dep[y] - dep[x];
    for (int i = 18; i >= 0; i--) {
      if ((de >> i) & 1) y = fa[y][i];
    }
    if (x == y) return x;
    for (int i = 18; i >= 0; i--) {
      if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
    }
    return fa[x][0];
  }
  void SOLVE() {
    int pd, u;
    scanf("%d%d", &pd, &u);
    if (pd == 1) {
      long long v;
      scanf("%lld", &v);
      if (v < 0) v = -v;
      long long now = v - a[u];
      a[u] = v;
      UPD(dfn[u], now);
      UPD(las[u] + 1, -now);
    }
    if (pd == 2) {
      int v;
      scanf("%d", &v);
      long long ans = VAL(u) + VAL(v) - 2 * VAL(LCA(u, v)) + a[LCA(u, v)];
      ans *= 2;
      ans -= a[u] + a[v];
      printf("%lld\n", ans);
    }
  }
  int main() {
    READ();
    INIT();
    while (T--) SOLVE();
    return 0;
  }
  