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 ll = long long;
  using std::cin;
  using std::cout;
  using std::endl;
  std::mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
  template <class T>
  inline bool chmax(T &a, T b) {
    if (a < b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  inline bool chmin(T &a, T b) {
    if (a > b) {
      a = b;
      return 1;
    }
    return 0;
  }
  const int inf = (int)1e9 + 7;
  const long long INF = 1LL << 60;
  namespace KKT89 {
  struct HLD {
    std::vector<std::vector<int>> g;
    std::vector<int> sz, in, out, head, par;
    HLD() {}
    HLD(std::vector<std::vector<int>> &_g)
        : g(_g),
          sz(_g.size()),
          in(_g.size()),
          out(_g.size()),
          head(_g.size()),
          par(_g.size()) {
      build();
    }
    void dfs_sz(int cur, int pre) {
      sz[cur] = 1;
      for (int &next : g[cur]) {
        if (next == pre) continue;
        dfs_sz(next, cur);
        par[next] = cur;
        sz[cur] += sz[next];
        if (sz[next] > sz[g[cur][0]]) std::swap(next, g[cur][0]);
      }
    }
    void dfs_hld(int cur, int pre, int &idx) {
      in[cur] = idx++;
      for (int &next : g[cur]) {
        if (next == pre) continue;
        if (next == g[cur][0])
          head[next] = head[cur];
        else
          head[next] = next;
        dfs_hld(next, cur, idx);
      }
      out[cur] = idx;
    }
    void build() {
      dfs_sz(0, -1);
      int idx = 0;
      dfs_hld(0, -1, idx);
    }
    void build(std::vector<std::vector<int>> &_g) {
      g = _g;
      sz.resize(g.size());
      in.resize(g.size());
      out.resize(g.size());
      head.resize(g.size());
      par.resize(g.size());
      dfs_sz(0, -1);
      int idx = 0;
      dfs_hld(0, -1, idx);
    }
    int lca(int u, int v) {
      for (;; v = par[head[v]]) {
        if (in[u] > in[v]) std::swap(u, v);
        if (head[u] == head[v]) return u;
      }
    }
    std::vector<std::pair<int, int>> path_query(int u, int v, bool edge = false) {
      std::vector<std::pair<int, int>> ret;
      for (;; v = par[head[v]]) {
        if (in[u] > in[v]) std::swap(u, v);
        if (head[u] == head[v]) break;
        ret.emplace_back(in[head[v]], in[v] + 1);
      }
      ret.emplace_back(in[u] + edge, in[v] + 1);
      return ret;
    }
  };
  }  // namespace KKT89
  namespace KKT89 {
  template <typename T>
  struct BinaryIndexedTree {
    int n;
    std::vector<T> bit;
    BinaryIndexedTree() : n(0) {}
    BinaryIndexedTree(int _n) : n(_n) { bit = std::vector<T>(n + 1); }
    void add1(int idx, T val) {
      for (int i = idx; i <= n; i += i & -i) bit[i] += val;
    }
    T sum1(int idx) {
      T res = 0;
      for (int i = idx; i > 0; i -= i & -i) res += bit[i];
      return res;
    }
    void add(int idx, T val) { add1(idx + 1, val); }
    T sum(int left, int right) { return sum1(right) - sum1(left); }
    int lower_bound(T x) {
      int res = 0;
      int k = 1;
      while (2 * k <= n) k <<= 1;
      for (; k > 0; k >>= 1) {
        if (res + k <= n and bit[res + k] < x) {
          x -= bit[res + k];
          res += k;
        }
      }
      return res;
    }
  };
  }  // namespace KKT89
  void solve([[maybe_unused]] int CASE) {
    int n, kkt;
    cin >> n >> kkt;
    std::vector<ll> a(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    std::vector<std::vector<int>> g(n);
    for (int i = 1; i < n; ++i) {
      int s, t;
      cin >> s >> t;
      s -= 1, t -= 1;
      g[s].emplace_back(t);
      g[t].emplace_back(s);
    }
    KKT89::HLD hld(g);
    KKT89::BinaryIndexedTree<ll> bit(n);
    auto update = [&](const int idx, const ll val) {
      const int in_idx = hld.in[idx];
      const ll cur = bit.sum(in_idx, in_idx + 1);
      bit.add(in_idx, val - cur);
    };
    for (int i = 0; i < n; ++i) {
      update(i, std::abs(a[i]));
    }
    while (kkt--) {
      int t;
      cin >> t;
      if (t == 1) {
        int u;
        cin >> u;
        u -= 1;
        ll c;
        cin >> c;
        a[u] = c;
        update(u, std::abs(c));
      } else {
        int u, v;
        cin >> u >> v;
        u -= 1, v -= 1;
        if (u == v) {
          cout << 0 << "\n";
          continue;
        }
        ll res = 0;
        for (const auto &[l, r] : hld.path_query(u, v)) {
          res += bit.sum(l, r);
        }
        res *= 2;
        res -= std::abs(a[u]);
        res -= std::abs(a[v]);
        cout << res << "\n";
      }
    }
  }
  int main() {
    std::cin.tie(nullptr);
    std::ios::sync_with_stdio(false);
    int kkt = 1;
    for (int jupi = 1; jupi <= kkt; jupi++) solve(jupi);
    return 0;
  }
  