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 MAXN = 100005;
  long long tree[MAXN];
  vector<int> g[MAXN];
  int curtime, ltime[MAXN], rtime[MAXN], par[20][MAXN], depth[MAXN];
  void dfs(int u, int parent = -1) {
    par[0][u] = parent;
    for (int j = 1; j < 20; j++) {
      if (par[j - 1][u] == -1)
        par[j][u] = -1;
      else
        par[j][u] = par[j - 1][par[j - 1][u]];
    }
    ltime[u] = curtime++;
    rtime[u] = ltime[u];
    for (int v : g[u]) {
      if (v == parent) continue;
      depth[v] = depth[u] + 1;
      dfs(v, u);
      rtime[u] = rtime[v];
    }
  }
  int lca(int u, int v) {
    if (depth[u] < depth[v]) swap(u, v);
    int d = depth[u] - depth[v];
    for (int j = 0; j < 20; j++)
      if (d & (1 << j)) {
        u = par[j][u];
      }
    if (u == v) return u;
    for (int j = 19; j >= 0; j--)
      if (par[j][u] != par[j][v]) {
        u = par[j][u];
        v = par[j][v];
      }
    return par[0][u];
  }
  void add(int u, long long val) {
    u++;
    while (u < MAXN) {
      tree[u] += val;
      u += u & (-u);
    }
  }
  long long sum(int u) {
    u++;
    long long ret = 0;
    while (u > 0) {
      ret += tree[u];
      u -= u & (-u);
    }
    return ret;
  }
  int main() {
    int n, q;
    scanf("%d %d", &n, &q);
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      a[i] = abs(a[i]);
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      u--, v--;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(0, -1);
    for (int i = 0; i < n; i++) {
      add(ltime[i], a[i]);
      add(rtime[i] + 1, -a[i]);
    }
    auto calc = [&](int u, int v) {
      long long cur = sum(ltime[u]) + sum(ltime[v]);
      int l = lca(u, v);
      cur -= sum(ltime[l]);
      if (par[0][l] != -1) cur -= sum(ltime[par[0][l]]);
      return cur;
    };
    while (q--) {
      int type;
      scanf("%d", &type);
      if (type == 1) {
        int u, c;
        scanf("%d %d", &u, &c);
        u--;
        c = abs(c);
        add(ltime[u], c - a[u]);
        add(rtime[u] + 1, a[u] - c);
        a[u] = c;
      } else {
        int u, v;
        scanf("%d %d", &u, &v);
        u--, v--;
        if (u == v) {
          printf("0\n");
          continue;
        }
        long long cur = 2LL * calc(u, v);
        cur -= sum(ltime[u]);
        if (par[0][u] != -1) cur += sum(ltime[par[0][u]]);
        cur -= sum(ltime[v]);
        if (par[0][v] != -1) cur += sum(ltime[par[0][v]]);
        printf("%lld\n", cur);
      }
    }
    return 0;
  }
  