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 MOD = 998244353;
  const long long lim = 1e18;
  inline int rd() {
    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 * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  const int N = 2e5 + 10;
  vector<int> g[N];
  int f[25][N], d[N];
  void dfs(int p, int fa) {
    f[0][p] = fa;
    d[p] = d[fa] + 1;
    for (int i = 1; i <= 20; ++i) f[i][p] = f[i - 1][f[i - 1][p]];
    for (int x : g[p])
      if (x != fa) dfs(x, p);
  }
  int lca(int x, int y) {
    if (d[x] < d[y]) swap(x, y);
    for (int i = 20; ~i; --i)
      if (d[x] - d[y] >> i & 1) x = f[i][x];
    if (x == y) return x;
    for (int i = 20; ~i; --i)
      if (f[i][x] != f[i][y]) x = f[i][x], y = f[i][y];
    return f[0][x];
  }
  int dfn[N], len = 0, pos[N];
  long long cnt[N];
  int n, q;
  long long C[N];
  int lowbit(int x) { return x & (-x); }
  void update(int x, long long num) {
    while (x < N) {
      C[x] += num;
      x += lowbit(x);
    }
  }
  long long query(int x) {
    long long num = 0;
    while (x) {
      num += C[x];
      x -= lowbit(x);
    }
    return num;
  }
  void dfs2(int u, int fa) {
    dfn[u] = ++len;
    for (auto to : g[u]) {
      if (to == fa) continue;
      dfs2(to, u);
    }
    pos[u] = len + 1;
  }
  int main() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &cnt[i]);
      cnt[i] = abs(cnt[i]);
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(1, 0);
    dfs2(1, 0);
    for (int i = 1; i <= n; i++) {
      update(dfn[i], cnt[i]);
      update(pos[i], -cnt[i]);
    }
    while (q--) {
      int op, x, y;
      scanf("%d%d%d", &op, &x, &y);
      if (op == 1) {
        y = abs(y);
        update(dfn[x], y - cnt[x]);
        update(pos[x], -(y - cnt[x]));
        cnt[x] = y;
      } else {
        long long ans;
        ans = (query(dfn[x]) + query(dfn[y]) - query(dfn[lca(x, y)]) -
               query(dfn[f[0][lca(x, y)]])) *
                  2ll -
              cnt[x] - cnt[y];
        printf("%lld\n", ans);
      }
    }
  }
  