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;
  long long int tree[4 * 100005];
  void update(int n, int st, int en, int idx, int val) {
    if (st == en) {
      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] = tree[lc] + tree[rc];
  }
  long long int query(int n, int st, int en, int l, int r) {
    if (r < st || en < l) return 0;
    if (l <= st && en <= r) return tree[n];
    int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
    long long int q1 = query(lc, st, mid, l, r);
    long long int q2 = query(rc, mid + 1, en, l, r);
    return q1 + q2;
  }
  int n;
  int par[100005];
  int szz[100005];
  vector<int> v[100005];
  int depth[100005];
  int nxt[100005];
  void dfs(int n, int pp = 0) {
    par[n] = pp;
    szz[n] = 1;
    for (int i = 0; i < v[n].size(); i++) {
      int vv = v[n][i];
      if (vv == pp) continue;
      depth[vv] = depth[n] + 1;
      dfs(vv, n);
      szz[n] += szz[vv];
      if (nxt[n] == -1 || szz[vv] > szz[nxt[n]]) nxt[n] = vv;
    }
  }
  int chain[100005];
  int num[100005];
  int csz[100005];
  int top[100005];
  int cnt = 1;
  int all;
  void hld(int n, int pp = -1) {
    chain[n] = cnt - 1;
    num[n] = all++;
    if (!csz[cnt - 1]) top[cnt - 1] = n;
    csz[cnt - 1]++;
    if (nxt[n] != -1) hld(nxt[n], n);
    for (int i = 0; i < v[n].size(); i++) {
      int vv = v[n][i];
      if (vv == pp || vv == nxt[n]) continue;
      cnt++;
      hld(vv, n);
    }
  }
  long long int tot[100005];
  long long int go(int a, int b) {
    long long int ans = 0;
    while (chain[a] != chain[b]) {
      if (depth[top[chain[a]]] < depth[top[chain[b]]]) swap(a, b);
      int start = top[chain[a]];
      if (num[a] == num[start] + csz[chain[a]] - 1)
        ans += tot[chain[a]];
      else
        ans += query(1, 1, n, num[start], num[a]);
      a = par[start];
    }
    if (depth[a] > depth[b]) swap(a, b);
    ans += query(1, 1, n, num[a], num[b]);
    return ans;
  }
  void modify(int node, int val) {
    update(1, 1, n, num[node], val);
    int st = num[top[chain[node]]];
    int en = st + csz[chain[node]] - 1;
    tot[chain[node]] = query(1, 1, n, st, en);
  }
  void ini() {
    memset(nxt, -1, sizeof(nxt));
    memset(v, NULL, sizeof(v));
    memset(csz, 0, sizeof(csz));
    cnt = 2;
    all = 1;
  }
  int cost[100005];
  int main() {
    int t, a, b, q, op;
    t = 1;
    while (t--) {
      ini();
      scanf("%d %d", &n, &q);
      for (int i = 1; i < n + 1; i++) {
        scanf("%d", &cost[i]);
        cost[i] = abs(cost[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);
      hld(1);
      for (int i = 1; i < n + 1; i++) modify(i, cost[i]);
      while (q--) {
        scanf("%d %d %d", &op, &a, &b);
        if (op == 1) {
          cost[a] = abs(b);
          modify(a, cost[a]);
        } else
          printf("%lld\n", 2 * go(a, b) - cost[a] - cost[b]);
      }
    }
    return 0;
  }
  