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;
  const long long int mod = 1000000007;
  long long int n, q;
  vector<long long int> a;
  vector<long long int> graph[100001];
  long long int parent[20][100001], sizes[100001], seg[4 * 100001], depth[100001],
      lazy[4 * 100001], mapping[100001], till[100001];
  void update(long long int v, long long int from, long long int to,
              long long int l, long long int r, long long int del) {
    if (lazy[v]) {
      seg[v] += (to - from + 1) * lazy[v];
      if (from != to) lazy[2 * v] += lazy[v], lazy[2 * v + 1] += lazy[v];
      lazy[v] = 0;
    }
    if (from > r || l > to) return;
    if (l <= from && to <= r) {
      seg[v] += (to - from + 1) * del;
      if (from != to) lazy[2 * v] += del, lazy[2 * v + 1] += del;
      return;
    }
    long long int mid = (from + to) / 2;
    update(2 * v, from, mid, l, r, del);
    update(2 * v + 1, mid + 1, to, l, r, del);
    seg[v] = seg[2 * v] + seg[2 * v + 1];
  }
  long long int query(long long int v, long long int from, long long int to,
                      long long int l, long long int r) {
    if (lazy[v]) {
      seg[v] += (to - from + 1) * lazy[v];
      if (from != to) lazy[2 * v] += lazy[v], lazy[2 * v + 1] += lazy[v];
      lazy[v] = 0;
    }
    if (from > r || l > to) return 0;
    if (l <= from && to <= r) return seg[v];
    long long int mid = (from + to) / 2;
    return query(2 * v, from, mid, l, r) + query(2 * v + 1, mid + 1, to, l, r);
  }
  void dfs(long long int node, long long int par, long long int &i) {
    mapping[node] = i;
    parent[0][node] = par;
    depth[node] = depth[par] + 1;
    sizes[node] = a[node];
    for (long long int e : graph[node]) {
      if (e == par) continue;
      dfs(e, node, ++i);
      sizes[node] += sizes[e];
    }
    till[node] = i;
    update(1, 1, n, mapping[node], till[node], a[node]);
  }
  long long int lca(long long int a, long long int b) {
    if (depth[a] < depth[b]) swap(a, b);
    long long int del = depth[a] - depth[b];
    long long int i = 19;
    while (del) {
      if (del & (1ll << i)) {
        del -= (1ll << i);
        a = parent[i][a];
      }
      i -= 1;
    }
    if (a == b) return a;
    for (long long int i = 19; i >= 0; i--) {
      if (parent[i][a] == parent[i][b]) continue;
      a = parent[i][a];
      b = parent[i][b];
    }
    return parent[0][a];
  }
  void solve() {
    cin >> n >> q;
    a.resize(n + 1);
    for (long long int i = 1; i <= n; i += 1) cin >> a[i], a[i] = abs(a[i]);
    long long int ch, x, y;
    for (long long int i = 1; i <= n - 1; i += 1) {
      cin >> x >> y;
      graph[x].push_back(y);
      graph[y].push_back(x);
    }
    depth[0] = -1;
    long long int val = 1;
    dfs(1, 0, val);
    for (long long int gen = 1; gen <= 19; gen += 1) {
      for (long long int node = 1; node <= n; node += 1)
        parent[gen][node] = parent[gen - 1][parent[gen - 1][node]];
    }
    while (q--) {
      cin >> ch >> x >> y;
      if (ch == 1) {
        y = abs(y);
        long long int del = y - a[x];
        update(1, 1, n, mapping[x], till[x], del);
        a[x] = y;
      } else {
        long long int lc = lca(x, y);
        long long int ans = 0;
        if (lc == y) swap(x, y);
        if (lc == x) {
          long long int px = parent[0][x];
          ans = query(1, 1, n, mapping[y], mapping[y]) -
                query(1, 1, n, mapping[px], mapping[px]);
        } else {
          ans = query(1, 1, n, mapping[y], mapping[y]) +
                query(1, 1, n, mapping[x], mapping[x]) -
                2 * query(1, 1, n, mapping[lc], mapping[lc]) + a[lc];
        }
        ans *= 2;
        ans -= a[x] + a[y];
        cout << ans << "\n";
      }
    }
  }
  int main() { solve(); }
  