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 maxn = 1e5 + 10;
  namespace Tree {
  int f[maxn], top[maxn], son[maxn], sz[maxn], id[maxn], rnk[maxn], depth[maxn];
  int tot;
  std::vector<int> vec[maxn];
  int n, q;
  void dfs1(int u) {
    sz[u] = 1;
    depth[u] = depth[f[u]] + 1;
    for (const auto &v : vec[u]) {
      if (v == f[u]) continue;
      f[v] = u;
      dfs1(v);
      sz[u] += sz[v];
      if (sz[v] > sz[son[u]]) {
        son[u] = v;
      }
    }
  }
  void dfs2(int u, int tp) {
    id[u] = ++tot;
    rnk[tot] = u;
    top[u] = tp;
    if (son[u]) {
      dfs2(son[u], tp);
    }
    for (const auto &v : vec[u]) {
      if (v == f[u] || v == son[u]) continue;
      dfs2(v, v);
    }
  }
  long long sum[maxn << 2];
  void pushup(int o) { sum[o] = sum[o << 1] + sum[o << 1 | 1]; }
  void update(int o, int L, int R, int pos, int v) {
    if (L == R) {
      sum[o] = v;
      return;
    }
    int mid = (L + R) / 2;
    if (pos <= mid)
      update(o << 1, L, mid, pos, v);
    else
      update(o << 1 | 1, mid + 1, R, pos, v);
    pushup(o);
  }
  long long query(int o, int L, int R, int l, int r) {
    if (l <= L && R <= r) {
      return sum[o];
    }
    int mid = (L + R) / 2;
    long long ret = 0;
    if (l <= mid) ret += query(o << 1, L, mid, l, r);
    if (r > mid) ret += query(o << 1 | 1, mid + 1, R, l, r);
    return ret;
  }
  long long query(int x, int y) {
    long long ret = 0;
    while (top[x] != top[y]) {
      if (depth[top[x]] < depth[top[y]]) {
        std::swap(x, y);
      }
      ret += query(1, 1, n, id[top[x]], id[x]);
      x = f[top[x]];
    }
    if (depth[x] > depth[y]) {
      std::swap(x, y);
    }
    ret += query(1, 1, n, id[x], id[y]);
    return ret;
  }
  };  // namespace Tree
  using namespace Tree;
  int arr[maxn];
  int main() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &arr[i]);
      arr[i] = std::abs(arr[i]);
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      vec[u].emplace_back(v);
      vec[v].emplace_back(u);
    }
    dfs1(1);
    dfs2(1, 1);
    for (int i = 1; i <= n; i++) {
      update(1, 1, n, id[i], arr[i]);
    }
    while (q--) {
      int op, u, v;
      scanf("%d%d%d", &op, &u, &v);
      v = std::abs(v);
      if (op == 1) {
        arr[u] = v;
        update(1, 1, n, id[u], v);
      } else {
        long long ans = query(u, v);
        ans *= 2;
        ans -= arr[u] + arr[v];
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  