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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  char _buf[100000], *_p1 = _buf, *_p2 = _buf;
  inline int gi() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 3) + (x << 1) + (ch ^ 48);
      ch = getchar();
    }
    return (f == 1) ? x : -x;
  }
  const int maxn = 1e5 + 5;
  int c[maxn], n, q;
  vector<int> e[maxn];
  int dfn[maxn], tot;
  struct BIT {
    long long c[maxn];
    inline void update(int x, int v) {
      while (x <= n) c[x] += v, x += x & -x;
    }
    inline long long query(int x) {
      long long ret = 0;
      while (x) ret += c[x], x -= x & -x;
      return ret;
    }
  } tree;
  inline void init() {}
  inline void input() {
    n = gi();
    q = gi();
    for (int i = 1; i <= n; ++i) c[i] = gi(), c[i] = abs(c[i]);
    for (int i = 1; i <= n - 1; ++i) {
      int u = gi(), v = gi();
      e[u].push_back(v), e[v].push_back(u);
    }
  }
  int fa[maxn][20], Log[maxn], dep[maxn], low[maxn];
  inline void dfs(int u) {
    dfn[u] = ++tot;
    dep[u] = dep[fa[u][0]] + 1;
    for (int i = 1; i <= Log[dep[u]]; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
    for (int v : e[u]) {
      if (v == fa[u][0]) continue;
      fa[v][0] = u;
      dfs(v);
    }
    low[u] = tot;
  }
  inline int LCA(int u, int v) {
    if (dep[u] > dep[v]) swap(u, v);
    for (int i = Log[dep[v]]; i >= 0; --i)
      if (dep[fa[v][i]] >= dep[u]) v = fa[v][i];
    if (u == v) return u;
    for (int i = Log[dep[v]]; i >= 0; --i)
      if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
    return fa[u][0];
  }
  inline void solve() {
    for (int i = 2; i <= n; ++i) Log[i] = Log[i >> 1] + 1;
    dfs(1);
    for (int i = 1; i <= n; ++i)
      tree.update(dfn[i], c[i]), tree.update(low[i] + 1, -c[i]);
    for (int i = 1; i <= q; ++i) {
      int opt = gi();
      int u = gi(), v = gi();
      if (opt == 1) {
        v = abs(v);
        tree.update(dfn[u], v - c[u]);
        tree.update(low[u] + 1, c[u] - v);
        c[u] = v;
      } else {
        long long ans = 0;
        int lca = LCA(u, v);
        ans += 2 * tree.query(dfn[u]) + 2 * tree.query(dfn[v]) -
               2 * tree.query(dfn[lca]) - 2 * tree.query(dfn[fa[lca][0]]);
        ans -= c[u] + c[v];
        printf("%lld\n", ans);
      }
    }
  }
  int main() {
    int T = 1;
    while (T--) {
      init();
      input();
      solve();
    }
    return 0;
  }
  