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 double eps = 1e-9;
  const long long INFll = 4e18;
  const long long INFi = 1e9 + 63;
  const double pi = acos(-1);
  const int maxn = 6e5;
  const int maxd = 16;
  const long long mod = 998244353;
  const int sqrtN = 300;
  int caseNo = 1;
  void E() {}
  template <class A, class... B>
  void E(A $, B... _) {
    cerr << ' ' << $;
    E(_...);
  }
  struct rmq_t {
    using val_t = long long;
    const static int size = 2e5;
    val_t mas[size << 1];
    static long long op(long long a, long long b) { return a + b; }
    void modify(int ind, val_t val) {
      ind += size;
      mas[ind] = val;
      for (ind >>= 1; ind > 0; ind >>= 1)
        mas[ind] = op(mas[ind << 1], mas[(ind << 1) + 1]);
    }
    val_t query(int l, int r) {
      if (l > r) return 0;
      l += size;
      r += size;
      val_t ans = mas[l];
      if (l < r) ans = op(ans, mas[r]);
      for (; l < r; l >>= 1, r >>= 1) {
        if ((l & 1) == 0 && (l + 1) < r) ans = op(ans, mas[l + 1]);
        if ((r & 1) == 1 && (r - 1) > l) ans = op(ans, mas[r - 1]);
      }
      return ans;
    }
  };
  rmq_t rmq;
  vector<int> g[maxn];
  namespace {
  const int maxn = 2e5;
  int par[maxn];
  int sz_sub[maxn];
  int mx[maxn];
  int pos[maxn];
  int root[maxn];
  int depth[maxn];
  void dfs_all(int v, int d = 0) {
    depth[v] = d;
    sz_sub[v] = 1;
    int ind = -1, mai = -1;
    for (auto it : g[v]) {
      if (par[v] != it) {
        par[it] = v;
        dfs_all(it, d + 1);
        sz_sub[v] += sz_sub[it];
        if (sz_sub[it] > mai) {
          mai = sz_sub[it];
          ind = it;
        }
      }
    }
    mx[v] = ind;
  }
  void build_array() {
    int cpos = 0;
    for (int i = 0; i < maxn; ++i) {
      if (sz_sub[i] && (par[i] == -1 || mx[par[i]] != i)) {
        for (int v = i; v != -1; v = mx[v]) {
          pos[v] = cpos++;
          root[v] = i;
        }
      }
    }
  }
  long long get(int u, int v) {
    long long ans = 0;
    while (root[u] != root[v]) {
      if (depth[root[u]] < depth[root[v]]) swap(u, v);
      ans = rmq_t::op(ans, rmq.query(pos[root[u]], pos[u]));
      u = par[root[u]];
    }
    if (depth[u] < depth[v]) swap(u, v);
    ans = rmq_t::op(ans, rmq.query(pos[v], pos[u]));
    return ans;
  }
  }  // namespace
  void solve() {
    int n, q;
    cin >> n >> q;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (int i = 0; i < n - 1; i++) {
      int s, t;
      cin >> s >> t;
      s--;
      t--;
      g[s].push_back(t);
      g[t].push_back(s);
    }
    dfs_all(0);
    build_array();
    for (int i = 0; i < n; i++) {
      rmq.modify(pos[i], a[i]);
    }
    for (int i = 0; i < q; i++) {
      int type;
      cin >> type;
      if (type == 1) {
        int u, c;
        cin >> u >> c;
        u--;
        rmq.modify(pos[u], abs(c));
      } else {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        long long test = 2 * get(u, v);
        test -= get(u, u);
        test -= get(v, v);
        cout << test << "\n";
      }
    }
  }
  int main(int argc, char** argv) {
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(false);
    cout << setprecision(15) << fixed;
    int t = 1;
    clock_t start = clock();
    for (int i = 0; i < t; i++) solve();
    return 0;
  }
  