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;
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
  const long long N = 1e5 + 10;
  vector<long long> g[N];
  vector<long long> d(N);
  vector<long long> arr;
  vector<long long> in(N), out(N);
  long long po[20][N], t[4 * N], add[4 * N];
  void dfs(long long v, long long p) {
    in[v] = arr.size();
    arr.push_back(v);
    for (auto to : g[v]) {
      if (to != p) {
        d[to] = d[v] + 1;
        po[0][to] = v;
        dfs(to, v);
      }
    }
    out[v] = (long long)(arr.size()) - 1;
  }
  long long lca(long long a, long long b) {
    if (d[a] < d[b]) {
      swap(a, b);
    }
    for (long long i = 19; i >= 0; i--) {
      if (d[po[i][a]] >= d[b]) {
        a = po[i][a];
      }
    }
    if (a == b) return a;
    for (long long i = 19; i >= 0; i--) {
      if (po[i][a] != po[i][b]) {
        a = po[i][a];
        b = po[i][b];
      }
    }
    return po[0][a];
  }
  void push(long long v) {
    t[v] += add[v];
    if (v * 2 < 4 * N) {
      add[v * 2] += add[v];
      add[v * 2 + 1] += add[v];
    }
    add[v] = 0;
  }
  void update(long long v, long long tl, long long tr, long long l, long long r,
              long long val) {
    push(v);
    if (l > r) return;
    if (tl == l && tr == r) {
      add[v] += val;
      push(v);
      return;
    }
    long long tm = (tl + tr) / 2;
    update(2 * v, tl, tm, l, min(r, tm), val),
        update(2 * v + 1, tm + 1, tr, max(l, tm + 1), r, val);
  }
  long long get(long long v, long long l, long long r, long long pos) {
    push(v);
    if (l == r) {
      return t[v];
    }
    long long m = (l + r) / 2;
    if (pos <= m)
      return get(2 * v, l, m, pos);
    else
      return get(2 * v + 1, m + 1, r, pos);
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long n, q;
    cin >> n >> q;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      a--, b--;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    dfs(0, -1);
    for (long long i = 1; i < 20; i++) {
      for (long long j = 0; j < n; j++) {
        po[i][j] = po[i - 1][po[i - 1][j]];
      }
    }
    for (long long i = 0; i < n; i++) {
      update(1, 0, n - 1, in[i], out[i], a[i]);
    }
    for (long long i = 0; i < q; i++) {
      long long t;
      cin >> t;
      if (t == 1) {
        long long v, c;
        cin >> v >> c;
        v--;
        update(1, 0, n - 1, in[v], out[v], -a[v]);
        a[v] = abs(c);
        update(1, 0, n - 1, in[v], out[v], a[v]);
      } else {
        long long u, v;
        cin >> u >> v;
        u--, v--;
        if (u == v) {
          cout << 0 << endl;
          continue;
        }
        long long g = lca(u, v), ans = 0;
        if (u != g && po[0][u] != g) {
          ans += (get(1, 0, n - 1, in[po[0][u]]) - get(1, 0, n - 1, in[g])) * 2;
        }
        if (v != g && po[0][v] != g) {
          ans += (get(1, 0, n - 1, in[po[0][v]]) - get(1, 0, n - 1, in[g])) * 2;
        }
        set<long long> st{u, v, g};
        for (auto to : st) {
          ans += a[to];
        }
        if (u != g && v != g) {
          ans += a[g];
        }
        cout << ans << endl;
      }
    }
    return 0;
  }
  