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 warning(disable : 4996)
  #pragma comment(linker, "/STACK:16777216")
  using namespace std;
  const int INF = 1000000000 + 1e8;
  const long long LINF = 2000000000000000000;
  int n;
  vector<int> a, b, sz, pos, head, depth, p;
  vector<vector<int>> g;
  int cur_pos = 0;
  const int N = 1e5 + 1010;
  long long t[4 * N];
  void build(int v, int vl, int vr) {
    if (vl == vr) {
      t[v] = {a[vl]};
      return;
    }
    int vm = (vl + vr) / 2;
    build(2 * v + 1, vl, vm);
    build(2 * v + 2, vm + 1, vr);
    t[v] = t[2 * v + 1] + t[2 * v + 2];
  }
  long long query(int v, int vl, int vr, int l, int r) {
    if (r < vl || vr < l) return 0;
    if (vl >= l && vr <= r) return t[v];
    int vm = (vl + vr) / 2;
    return query(2 * v + 1, vl, vm, l, r) + query(2 * v + 2, vm + 1, vr, l, r);
  }
  void modify(int v, int vl, int vr, int pos, int val) {
    if (vl == vr) {
      t[v] = val;
      return;
    }
    int vm = (vl + vr) / 2;
    if (pos <= vm) {
      modify(2 * v + 1, vl, vm, pos, val);
    } else
      modify(2 * v + 2, vm + 1, vr, pos, val);
    t[v] = t[2 * v + 1] + t[2 * v + 2];
  }
  long long tree_query(int l, int r) { return query(0, 0, n - 1, l, r); }
  void dfs_sz(int v, int pr = -1) {
    sz[v] = 1;
    p[v] = pr;
    for (auto &to : g[v])
      if (to != pr) {
        depth[to] = depth[v] + 1;
        dfs_sz(to, v);
        sz[v] += sz[to];
        if (sz[to] > sz[g[v][0]]) swap(g[v][0], to);
      }
  }
  void dfs_hld(int v) {
    pos[v] = cur_pos++;
    for (auto to : g[v])
      if (to != p[v]) {
        head[to] = (to == g[v][0] ? head[v] : to);
        dfs_hld(to);
      }
  }
  long long query(int u, int v) {
    long long ans = 0;
    for (; head[u] != head[v]; v = p[head[v]]) {
      if (depth[head[u]] > depth[head[v]]) swap(u, v);
      ans += tree_query(pos[head[v]], pos[v]);
    }
    if (depth[u] > depth[v]) swap(u, v);
    ans += tree_query(pos[u], pos[v]);
    return ans;
  }
  void tmodify(int v, int x) { modify(0, 0, n - 1, pos[v], x); }
  void init() {
    sz.resize(n);
    pos.resize(n);
    head.resize(n);
    depth.resize(n);
    p.resize(n);
    b.resize(n);
    dfs_sz(0);
    dfs_hld(0);
    for (int i = 0; i < n; i++) b[pos[i]] = a[i];
    a = b;
    build(0, 0, n - 1);
  }
  void solve() {
    int q;
    cin >> n >> q;
    g.resize(n);
    a.resize(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      u--, v--;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    init();
    while (q--) {
      int type;
      cin >> type;
      if (type == 1) {
        int v, c;
        cin >> v >> c;
        v--;
        c = abs(c);
        tmodify(v, c);
      } else {
        int u, v;
        cin >> u >> v;
        u--, v--;
        long long sum = 2 * query(u, v) - query(u, u) - query(v, v);
        cout << sum << '\n';
      }
    }
  }
  int main() {
    srand(time(0));
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tst = 1;
    while (tst--) {
      solve();
    }
  }
  