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;
  struct SegmentTreeNode {
    long long int valueSum;
    int height;
    int node;
    SegmentTreeNode() {
      valueSum = 0;
      height = INT_MAX / 2;
    }
    SegmentTreeNode(long long int valueSum, int height, int node) {
      this->valueSum = valueSum;
      this->height = height;
      this->node = node;
    }
  };
  SegmentTreeNode merge(SegmentTreeNode a, SegmentTreeNode b) {
    a.valueSum += b.valueSum;
    if (a.height >= b.height) a.node = b.node;
    a.height = min(a.height, b.height);
    return a;
  }
  SegmentTreeNode tree[4 * 3 * 100005];
  void update(int n, int st, int en, int idx, SegmentTreeNode val) {
    if (st == en) {
      if (val.height == -1) val.height = tree[n].height;
      tree[n] = val;
      return;
    }
    int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
    if (idx <= mid)
      update(lc, st, mid, idx, val);
    else
      update(rc, mid + 1, en, idx, val);
    tree[n] = merge(tree[lc], tree[rc]);
  }
  SegmentTreeNode query(int n, int st, int en, int l, int r) {
    if (r < st || en < l) return SegmentTreeNode();
    if (l <= st && en <= r) return tree[n];
    int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
    auto q1 = query(lc, st, mid, l, r);
    auto q2 = query(rc, mid + 1, en, l, r);
    return merge(q1, q2);
  }
  int dfsTime;
  int starTime[3 * 100005];
  int endTime[3 * 100005];
  int value[3 * 100005];
  vector<int> v[3 * 100005];
  void dfs(int node, int height = 0, int par = 1) {
    starTime[node] = dfsTime++;
    update(1, 1, 3 * 100005, starTime[node], {value[node], height, node});
    for (auto child : v[node]) {
      if (child == par) continue;
      dfs(child, height + 1, node);
      update(1, 1, 3 * 100005, dfsTime, {0, height, node});
      dfsTime++;
    }
    endTime[node] = dfsTime++;
    update(1, 1, 3 * 100005, endTime[node], {-value[node], height, node});
  }
  void init() { dfsTime = 1; }
  int getLCA(int a, int b) {
    if (starTime[a] > starTime[b]) swap(a, b);
    SegmentTreeNode q = query(1, 1, 3 * 100005, starTime[a], starTime[b]);
    return q.node;
  }
  void update(int node, int newValue) {
    value[node] = newValue;
    update(1, 1, 3 * 100005, starTime[node], {value[node], -1, node});
    update(1, 1, 3 * 100005, endTime[node], {-value[node], -1, node});
  }
  long long int getPathSum(int a, int b) {
    int lca = getLCA(a, b);
    long long int aSum =
        query(1, 1, 3 * 100005, starTime[lca], starTime[a]).valueSum;
    long long int bSum =
        query(1, 1, 3 * 100005, starTime[lca], starTime[b]).valueSum;
    return aSum + bSum - value[lca];
  }
  int main() {
    init();
    int a, b, n, q;
    scanf("%d %d", &n, &q);
    for (int i = 1; i < n + 1; i++) {
      scanf("%d", &value[i]);
      value[i] = abs(value[i]);
    }
    for (int i = 0; i < n - 1; i++) {
      scanf("%d %d", &a, &b);
      v[a].push_back(b);
      v[b].push_back(a);
    }
    dfs(1);
    int op;
    while (q--) {
      scanf("%d %d %d", &op, &a, &b);
      if (op == 2)
        printf("%lld\n", 2 * getPathSum(a, b) - value[a] - value[b]);
      else
        update(a, abs(b));
    }
    return 0;
  }
  