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 N = 1e5 + 5;
  int n, dep[N], dfn[N], lst[N], cnt, f[N][25], eid, p[N], a[N];
  long long tr[N];
  struct edge {
    int v, nxt;
  } e[N << 1];
  inline void addedge(int u, int v) { e[++eid] = (edge){v, p[u]}, p[u] = eid; }
  inline int lowbit(int x) { return x & (-x); }
  inline void add(int x, long long d) {
    while (x <= n) tr[x] += d, x += lowbit(x);
  }
  inline void change(int l, int r, long long d) { add(l, d), add(r + 1, -d); }
  inline long long query(int x) {
    long long res = 0;
    while (x > 0) res += tr[x], x -= lowbit(x);
    return res;
  }
  inline void dfs(int u, int fa) {
    f[u][0] = fa, dep[u] = dep[fa] + 1, dfn[u] = ++cnt;
    for (int i = 1; i < 20 && f[u][i - 1]; i++) f[u][i] = f[f[u][i - 1]][i - 1];
    for (int i = p[u], v = e[i].v; i; v = e[i = e[i].nxt].v)
      if (v != fa) dfs(v, u);
    lst[u] = cnt;
  }
  inline int lca(int u, int v) {
    if (dep[u] < dep[v]) swap(u, v);
    for (int i = (19); i >= (0); i--)
      if (dep[f[u][i]] >= dep[v]) u = f[u][i];
    if (u == v) return u;
    for (int i = (19); i >= (0); i--)
      if (f[v][i] != f[u][i]) u = f[u][i], v = f[v][i];
    return f[u][0];
  }
  int main() {
    int q;
    scanf("%d%d", &n, &q);
    for (int i = (1); i <= (n); i++) scanf("%d", &a[i]), a[i] = max(a[i], -a[i]);
    for (int i = (1); i <= (n - 1); i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      addedge(u, v), addedge(v, u);
    }
    dfs(1, 0);
    for (int i = (1); i <= (n); i++) change(dfn[i], lst[i], a[i]);
    while (q--) {
      int opt, x, y;
      scanf("%d%d%d", &opt, &x, &y);
      if (opt == 1) {
        y = max(y, -y);
        change(dfn[x], lst[x], y - a[x]);
        a[x] = y;
      } else {
        int c = lca(x, y);
        long long w1 = query(dfn[x]) + query(dfn[y]);
        long long w2 = query(dfn[c]) + query(dfn[f[c][0]]);
        printf("%lld\n", (w1 - w2) * 2 - a[x] - a[y]);
      }
    }
    return 0;
  }
  