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 = 100000 + 9;
  struct BIT {
    int n;
    long long s[N];
    void init(int _n) { n = _n; }
    void upd(int p, long long k) {
      for (; p <= n; p += p & -p) s[p] += k;
    }
    long long sum(int p) {
      long long r = 0;
      for (; p; p -= p & -p) r += s[p];
      return r;
    }
    long long sum(int l, int r) { return sum(r) - sum(l - 1); }
  } bit;
  long long a[N], b[N];
  int tot, p[N], dep[N], sz[N], hc[N], top[N], dfn[N];
  vector<int> G[N];
  void dfs1(int u, int _p) {
    dep[u] = dep[p[u] = _p] + 1;
    sz[u] = 1;
    hc[u] = 0;
    for (int v : G[u])
      if (v != _p) {
        dfs1(v, u);
        sz[u] += sz[v];
        if (sz[v] > sz[hc[u]]) hc[u] = v;
      }
  }
  void dfs2(int u, int _top) {
    top[u] = _top;
    dfn[u] = ++tot;
    if (hc[u]) dfs2(hc[u], _top);
    for (int v : G[u])
      if (v != p[u] && v != hc[u]) dfs2(v, v);
  }
  void upd(int u, int c) {
    a[u] = c;
    if (u != 1) {
      long long pre = b[dfn[u]];
      b[dfn[u]] = max(abs(a[u] + a[p[u]]), abs(a[u] - a[p[u]]));
      bit.upd(dfn[u], b[dfn[u]] - pre);
    }
    if (hc[u]) {
      int v = hc[u];
      long long pre = b[dfn[v]];
      b[dfn[v]] = max(abs(a[v] + a[p[v]]), abs(a[v] - a[p[v]]));
      bit.upd(dfn[v], b[dfn[v]] - pre);
    }
  }
  long long query(int u, int v) {
    long long dis = 0;
    for (; top[u] != top[v]; u = p[top[u]]) {
      if (dep[top[u]] < dep[top[v]]) swap(u, v);
      dis += bit.sum(dfn[top[u]], dfn[u]);
      int w = top[u];
      dis += max(abs(a[w] + a[p[w]]), abs(a[w] - a[p[w]])) - b[dfn[w]];
    }
    if (dep[u] > dep[v]) swap(u, v);
    if (u != v) dis += bit.sum(dfn[u] + 1, dfn[v]);
    return dis;
  }
  int main() {
    int n, q;
    scanf("%d %d", &n, &q);
    for (int i = 1; i <= n; ++i) {
      scanf("%lld", a + 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);
    }
    dfs1(1, 0);
    dfs2(1, 1);
    bit.init(n);
    for (int i = 2; i <= n; ++i) {
      b[dfn[i]] = max(abs(a[i] + a[p[i]]), abs(a[i] - a[p[i]]));
      bit.upd(dfn[i], b[dfn[i]]);
    }
    for (int i = 1; i <= q; ++i) {
      int op;
      scanf("%d", &op);
      if (op == 1) {
        int u, c;
        scanf("%d %d", &u, &c);
        upd(u, c);
      } else if (op == 2) {
        int u, v;
        scanf("%d %d", &u, &v);
        long long ans = query(u, v);
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  