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;
  using ll = long long;
  using ull = unsigned long long;
  const int N = 1e5 + 100;
  const int LOG = 18;
  int cnt;
  bool used[N];
  ll a[N];
  int sz[N], comp[N], comp_sz[N], comp_pos[N], comp_prev[N], h[N];
  int p[LOG][N];
  vector<int> g[N];
  vector<ll> val[N], fen[N];
  void init() {}
  void dfs(int v) {
    used[v] = true;
    sz[v] = 1;
    for (int u : g[v]) {
      if (!used[u]) {
        p[0][u] = v;
        h[u] = h[v] + 1;
        dfs(u);
        sz[v] += sz[u];
      }
    }
  }
  void dfs2(int v, int c) {
    comp[v] = c;
    comp_pos[v] = comp_sz[c]++;
    pair<int, int> best(-1, -1);
    for (int u : g[v]) {
      if (p[0][u] == v) {
        best = max(best, make_pair(sz[u], u));
      }
    }
    if (best.second != -1) {
      dfs2(best.second, c);
    }
    for (int u : g[v]) {
      if (p[0][u] == v && u != best.second) {
        comp_prev[cnt++] = u;
        dfs2(u, cnt - 1);
      }
    }
  }
  void add(vector<ll>& f, int pos, ll val) {
    while (pos < ((int)(f).size())) {
      f[pos] += val;
      pos = (pos | (pos + 1));
    }
  }
  void set_val(int v, ll new_val) {
    add(fen[comp[v]], comp_pos[v], new_val - val[comp[v]][comp_pos[v]]);
    val[comp[v]][comp_pos[v]] = new_val;
  }
  int get_lca(int a, int b) {
    if (h[a] < h[b]) {
      swap(a, b);
    }
    for (int i = LOG - 1; i >= 0; --i) {
      if (p[i][a] != -1 && h[p[i][a]] >= h[b]) {
        a = p[i][a];
      }
    }
    if (a == b) {
      return a;
    }
    for (int i = LOG - 1; i >= 0; --i) {
      if (p[i][a] != p[i][b]) {
        a = p[i][a];
        b = p[i][b];
        assert(a != -1);
        assert(b != -1);
      }
    }
    return p[0][a];
  }
  ll get(const vector<ll>& f, int pos) {
    ll res = 0;
    while (pos >= 0) {
      res += f[pos];
      pos = (pos & (pos + 1)) - 1;
    }
    return res;
  }
  ll get(const vector<ll>& f, int l, int r) {
    assert(l <= r);
    return get(f, r) - get(f, l - 1);
  }
  ll get_sum(int a, int b) {
    assert(h[a] >= h[b]);
    ll res = 0;
    while (comp[a] != comp[b]) {
      res += get(fen[comp[a]], comp_pos[a]);
      a = p[0][comp_prev[comp[a]]];
    }
    assert(comp[a] == comp[b]);
    assert(comp_pos[b] <= comp_pos[a]);
    res += get(fen[comp[a]], comp_pos[b], comp_pos[a]);
    return res;
  }
  void solve() {
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; ++i) {
      scanf("%lld", a + i);
      a[i] = abs(a[i]);
    }
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      scanf("%d %d", &a, &b);
      --a;
      --b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    p[0][0] = -1;
    h[0] = 0;
    dfs(0);
    for (int i = 1; i < LOG; ++i) {
      for (int j = 0; j < n; ++j) {
        if (p[i - 1][j] == -1) {
          p[i][j] = -1;
        } else {
          p[i][j] = p[i - 1][p[i - 1][j]];
        }
      }
    }
    cnt = 1;
    comp_prev[0] = 0;
    dfs2(0, 0);
    for (int i = 0; i < cnt; ++i) {
      val[i].resize(comp_sz[i]);
      fen[i].resize(comp_sz[i]);
    }
    for (int i = 0; i < n; ++i) {
      set_val(i, a[i]);
    }
    for (int i = 0; i < m; ++i) {
      int t;
      scanf("%d", &t);
      if (t == 1) {
        int v;
        ll val;
        scanf("%d %lld", &v, &val);
        --v;
        a[v] = abs(val);
        set_val(v, a[v]);
      } else {
        assert(t == 2);
        int x, y;
        scanf("%d %d", &x, &y);
        --x;
        --y;
        int z = get_lca(x, y);
        ll res = 2 * (get_sum(x, z) + get_sum(y, z) - a[z]) - a[x] - a[y];
        printf("%lld\n", res);
      }
    }
  }
  int main() {
    init();
    int t = 1;
    for (int i = 0; i < t; ++i) {
      solve();
    }
    return 0;
  }
  