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 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long R = 7 + 1e9, R1 = 19491001, R2 = 236, NUMTESTCASE;
  const long long NN = 10 + 5e5, MM = 10 + 13e5, KK = 26, Inverse2 = 500000004;
  const double pi = acos(-1.0);
  int di[8] = {1, 0, -1, 0, 1, -1, 1, -1}, dj[8] = {0, 1, 0, -1, 1, -1, -1, 1};
  int Dsu[NN], n, q, u, v, x, Timer, Arr[NN];
  int Anc[NN][KK], Xor[NN], Depth[NN], Hash[NN];
  int Heavy[NN], Chain[NN], TotH, Head[NN], UpHead[NN];
  long long STree[MM], Lazy[MM], t;
  vector<int> G[NN];
  int Root(int u) { return Dsu[u] == u ? u : Dsu[u] = Root(Dsu[u]); }
  bool Merge(int u, int v) {
    u = Root(u);
    v = Root(v);
    if (u == v) return false;
    if (u > v) swap(u, v);
    Dsu[u] = v;
    return true;
  }
  void Hld(int Ver) {
    UpHead[Chain[Ver]] = Anc[Ver][0];
    if (Heavy[Anc[Ver][0]] == Ver)
      Hld(Head[Chain[Anc[Ver][0]] = Chain[Ver]] = Anc[Ver][0]);
  }
  int Dfs(int Ver, int Par) {
    Depth[Ver] = Depth[Par] + 1;
    Anc[Ver][0] = Par;
    for (int k = 1; k < KK; k++) Anc[Ver][k] = Anc[Anc[Ver][k - 1]][k - 1];
    int Size = 1, Temp;
    pair<int, int> Max = {1, Ver};
    for (auto u : G[Ver])
      if (u ^ Par) Max = max(Max, {Temp = Dfs(u, Ver), u}), Size += Temp;
    Heavy[Ver] = Max.second;
    if (Max.second == Ver) Heavy[Ver] = -1;
    return Size;
  }
  int Lca(int u, int v) {
    if (Depth[u] < Depth[v]) swap(u, v);
    for (int k = KK - 1; k >= 0; k--)
      if (Depth[Anc[u][k]] >= Depth[v]) u = Anc[u][k];
    if (u == v) return u;
    for (int k = KK - 1; k >= 0; k--)
      if (Anc[u][k] != Anc[v][k]) u = Anc[u][k], v = Anc[v][k];
    return Anc[u][0];
  }
  int Kth(int Ver, int k) {
    for (int j = 0; j < KK; j++)
      if (k >> j & 1) Ver = Anc[Ver][j];
    return Ver;
  }
  long long XXXX;
  void UpdateLazy(int Node, int Len, int p) {
    if (!Lazy[Node]) return;
    assert(Len == 0);
    STree[Node] = max(abs(Arr[p] + Arr[Anc[p][0]]), abs(Arr[p] - Arr[Anc[p][0]]));
    STree[Node] = XXXX;
    if (Len) Lazy[2 * Node] = Lazy[2 * Node + 1] = 1;
    Lazy[Node] = 0;
  }
  void UpdateSeg(int s, int e, int st = 1, int en = n, int Node = 1) {
    UpdateLazy(Node, en - st, st);
    if (s > en || e < st) return;
    if (s <= st && e >= en) {
      Lazy[Node] = 1;
      return UpdateLazy(Node, en - st, st);
    }
    UpdateSeg(s, e, st, (st + en) / 2, 2 * Node);
    UpdateSeg(s, e, (st + en) / 2 + 1, en, 2 * Node + 1);
    STree[Node] = STree[2 * Node] + STree[2 * Node + 1];
  }
  long long GetSeg(int s, int e, int st = 1, int en = n, int Node = 1) {
    UpdateLazy(Node, en - st, st);
    if (s > en || e < st) return 0;
    if (s <= st && e >= en) return STree[Node];
    return GetSeg(s, e, st, (st + en) / 2, 2 * Node) +
           GetSeg(s, e, (st + en) / 2 + 1, en, 2 * Node + 1);
  }
  long long GetSim(int u, int p) {
    if (Chain[u] == Chain[p]) return GetSeg(Hash[p], Hash[u]);
    return GetSim(u, Head[Chain[u]]) + GetSim(UpHead[Chain[u]], p);
  }
  long long GetPath(int u, int v) {
    int Lca_ = Lca(u, v);
    long long Res = 0;
    if (u ^ Lca_) Res += GetSim(u, Kth(u, Depth[u] - Depth[Lca_] - 1));
    if (v ^ Lca_) Res += GetSim(v, Kth(v, Depth[v] - Depth[Lca_] - 1));
    return Res;
  }
  void Update(int u) { UpdateSeg(Hash[u], Hash[u]); }
  void UpdateSim(int u, int p) {
    if (Chain[u] == Chain[p]) return UpdateSeg(Hash[p], Hash[u]);
    UpdateSim(UpHead[Chain[u]], p);
    UpdateSim(u, Head[Chain[u]]);
  }
  void UpdatePath(int u, int v) {
    int Lca_ = Lca(u, v);
    if (u ^ Lca_) UpdateSim(u, Kth(u, Depth[u] - Depth[Lca_] - 1));
    if (v ^ Lca_) UpdateSim(v, Kth(v, Depth[v] - Depth[Lca_] - 1));
  }
  int main() {
    cin >> n >> q;
    for (int i = (1); i <= (n); ++i) {
      Dsu[i] = i;
      scanf("%d", Arr + i);
    }
    for (int i = (2); i <= (n); ++i)
      scanf("%d%d", &u, &v), G[u].push_back(v), G[v].push_back(u);
    Dfs(1, 1);
    for (int u = (1); u <= (n); ++u)
      if (Heavy[u] == -1) {
        Hld(Head[Chain[u] = ++TotH] = u);
        for (int v = Head[TotH]; v + 1; v = Heavy[v]) Hash[v] = ++Timer;
      }
    for (int u = (1); u <= (n); ++u) {
      XXXX = max(abs(Arr[u] + Arr[Anc[u][0]]), abs(Arr[u] - Arr[Anc[u][0]]));
      Update(u);
    }
    for (int i = (1); i <= (q); ++i) {
      scanf("%d%d%d", &t, &u, &v);
      if (t == 2)
        printf("%lld\n", GetPath(v, u));
      else {
        Arr[u] = v;
        XXXX = max(abs(Arr[u] + Arr[Anc[u][0]]), abs(Arr[u] - Arr[Anc[u][0]]));
        Update(u);
        for (int p : G[u]) {
          if (p ^ Anc[u][0]) {
            XXXX =
                max(abs(Arr[p] + Arr[Anc[p][0]]), abs(Arr[p] - Arr[Anc[p][0]]));
            Update(p);
          }
        }
      }
    }
    return 0;
  }
  