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 namespace std::chrono;
  template <typename A>
  ostream& operator<<(ostream& cout, vector<A> const& v);
  template <typename A, typename B>
  ostream& operator<<(ostream& cout, pair<A, B> const& p) {
    return cout << "(" << p.first << ", " << p.second << ")";
  }
  template <typename A>
  ostream& operator<<(ostream& cout, vector<A> const& v) {
    cout << "[";
    for (int i = 0; i < v.size(); i++) {
      if (i) cout << ", ";
      cout << v[i];
    }
    return cout << "]";
  }
  template <typename A, typename B>
  istream& operator>>(istream& cin, pair<A, B>& p) {
    cin >> p.first;
    return cin >> p.second;
  }
  mt19937 rng(61378913);
  void usaco(string filename) {
    freopen((filename + ".in").c_str(), "r", stdin);
    freopen((filename + ".out").c_str(), "w", stdout);
  }
  const long double pi = 3.14159265358979323846;
  long long n, m, k, q, l, r, x, y, z;
  long long a[1000005];
  long long b[1000005];
  long long c[1000005];
  string second, t;
  long long ans = 0;
  template <int size, int lg, typename seg_t = long long>
  struct hld {
    vector<int> edges[size];
    int bigchild[size];
    int sz[size];
    int depth[size];
    int chain[size];
    int label[size], label_time;
    int par[size];
    int lca_lift[size][lg];
    seg_t seg_tree[4 * size];
    seg_t seg_lazy[4 * size];
    int N;
    inline seg_t seg_combine(seg_t a, seg_t b) { return a + b; }
    inline seg_t seg_lazy_apply(seg_t lazy_val, seg_t new_val) { return new_val; }
    inline seg_t seg_lazy_func(seg_t cur_val, seg_t lazy_val, int l, int r) {
      return lazy_val;
    }
    const seg_t seg_sentinel = 0;
    const seg_t seg_lazy_sentinel = -1;
    const seg_t seg_init_val = 0;
    seg_t seg_query_header(int l, int r) {
      return seg_query_rec(0, 0, N - 1, l, r);
    }
    seg_t seg_query_rec(int i, int tl, int tr, int ql, int qr) {
      seg_eval_lazy(i, tl, tr);
      if (ql <= tl && tr <= qr) return seg_tree[i];
      if (tl > tr || tr < ql || qr < tl) return seg_sentinel;
      int mid = (tl + tr) / 2;
      seg_t a = seg_query_rec(2 * i + 1, tl, mid, ql, qr);
      seg_t b = seg_query_rec(2 * i + 2, mid + 1, tr, ql, qr);
      return seg_combine(a, b);
    }
    void seg_update_header(int l, int r, seg_t v) {
      seg_update_rec(0, 0, N - 1, l, r, v);
    }
    seg_t seg_update_rec(int i, int tl, int tr, int ql, int qr, seg_t v) {
      seg_eval_lazy(i, tl, tr);
      if (tl > tr || tr < ql || qr < tl) return seg_tree[i];
      if (ql <= tl && tr <= qr) {
        seg_lazy[i] = seg_lazy_apply(seg_lazy[i], v);
        seg_eval_lazy(i, tl, tr);
        return seg_tree[i];
      }
      if (tl == tr) return seg_tree[i];
      int mid = (tl + tr) / 2;
      seg_t a = seg_update_rec(2 * i + 1, tl, mid, ql, qr, v);
      seg_t b = seg_update_rec(2 * i + 2, mid + 1, tr, ql, qr, v);
      return seg_tree[i] = seg_combine(a, b);
    }
    void seg_eval_lazy(int i, int l, int r) {
      if (seg_lazy[i] == seg_lazy_sentinel) return;
      seg_tree[i] = seg_lazy_func(seg_tree[i], seg_lazy[i], l, r);
      if (l != r) {
        seg_lazy[i * 2 + 1] = seg_lazy_apply(seg_lazy[i * 2 + 1], seg_lazy[i]);
        seg_lazy[i * 2 + 2] = seg_lazy_apply(seg_lazy[i * 2 + 2], seg_lazy[i]);
      }
      seg_lazy[i] = seg_lazy_sentinel;
    }
    void init_arrays(int n) {
      N = n;
      for (int i = 0; i < N; i++) {
        edges[i].clear();
        chain[i] = i;
      }
      for (int i = 0; i < 4 * N; i++) {
        seg_tree[i] = seg_init_val;
        seg_lazy[i] = seg_lazy_sentinel;
      }
    }
    void add_edge(int u, int v) {
      edges[u].push_back(v);
      edges[v].push_back(u);
    }
    void init_tree(seg_t* arr, int root = 0) {
      lca_dfs(root, -1);
      dfs_size(root, -1, 0);
      dfs_chains(root, -1);
      label_time = 0;
      dfs_labels(arr, root, -1);
    }
    void lca_dfs(int v, int par) {
      lca_lift[v][0] = par;
      for (int i = 1; i < lg; i++) {
        if (lca_lift[v][i - 1] == -1)
          lca_lift[v][i] = -1;
        else
          lca_lift[v][i] = lca_lift[lca_lift[v][i - 1]][i - 1];
      }
      for (int x : edges[v]) {
        if (x != par) {
          lca_dfs(x, v);
        }
      }
    }
    void dfs_size(int v, int p, int d) {
      sz[v] = 1;
      depth[v] = d;
      par[v] = p;
      int bigc = -1, bigv = -1;
      for (int x : edges[v]) {
        if (x != p) {
          dfs_size(x, v, d + 1);
          sz[v] += sz[x];
          if (sz[x] > bigv) {
            bigc = x;
            bigv = sz[x];
          }
        }
      }
      bigchild[v] = bigc;
    }
    void dfs_chains(int v, int p) {
      if (bigchild[v] != -1) {
        chain[bigchild[v]] = chain[v];
      }
      for (int x : edges[v]) {
        if (x != p) {
          dfs_chains(x, v);
        }
      }
    }
    void dfs_labels(seg_t* arr, int v, int p) {
      label[v] = label_time++;
      seg_update_header(label[v], label[v], arr[v]);
      if (bigchild[v] != -1) {
        dfs_labels(arr, bigchild[v], v);
      }
      for (int x : edges[v]) {
        if (x != p && x != bigchild[v]) {
          dfs_labels(arr, x, v);
        }
      }
    }
    int lca(int a, int b) {
      if (depth[a] < depth[b]) swap(a, b);
      int d = depth[a] - depth[b];
      int v = get_kth_ancestor(a, d);
      if (v == b) {
        return v;
      } else {
        for (int i = lg - 1; i >= 0; i--) {
          if (lca_lift[v][i] != lca_lift[b][i]) {
            v = lca_lift[v][i];
            b = lca_lift[b][i];
          }
        }
        return lca_lift[b][0];
      }
    }
    int get_kth_ancestor(int v, int k) {
      for (int i = lg - 1; i >= 0; i--) {
        if (v == -1) return v;
        if ((1 << i) <= k) {
          v = lca_lift[v][i];
          k -= (1 << i);
        }
      }
      return v;
    }
    seg_t query_chain(int v, int p) {
      seg_t val = seg_sentinel;
      while (depth[p] < depth[v]) {
        int top = chain[v];
        if (depth[top] <= depth[p]) {
          int diff = depth[v] - depth[p];
          top = get_kth_ancestor(v, diff - 1);
        }
        val = seg_combine(val, seg_query_header(label[top], label[v]));
        v = par[top];
      }
      return val;
    }
    seg_t query(int u, int v) {
      int lc = lca(u, v);
      seg_t val = seg_combine(query_chain(u, lc), query_chain(v, lc));
      return seg_combine(val, seg_query_header(label[lc], label[lc]));
    }
    void update_chain(int v, int p, seg_t val) {
      while (depth[p] < depth[v]) {
        int top = chain[v];
        if (depth[top] <= depth[p]) {
          int diff = depth[v] - depth[p];
          top = get_kth_ancestor(v, diff - 1);
        }
        seg_update_header(label[top], label[v], val);
        v = par[top];
      }
    }
    void update(int u, int v, seg_t val) {
      int lc = lca(u, v);
      update_chain(u, lc, val);
      update_chain(v, lc, val);
      seg_update_header(label[lc], label[lc], val);
    }
  };
  hld<100005, 20, long long> h;
  void solve(int tc) {
    cin >> n >> q;
    for (int ele = 0; ele < n; ele++) cin >> a[ele];
    ;
    for (int i = 0; i < n; i++) a[i] = abs(a[i]);
    h.init_arrays(n);
    for (long long i = 0; i < (n - 1); ++i) {
      cin >> x >> y;
      --x;
      --y;
      h.add_edge(x, y);
    }
    h.init_tree(a);
    for (long long i = 0; i < (q); ++i) {
      int t;
      cin >> t;
      if (t == 1) {
        int v;
        cin >> x >> v;
        --x;
        v = abs(v);
        a[x] = v;
        h.update(x, x, v);
      } else {
        cin >> x >> y;
        --x;
        --y;
        long long res = h.query(x, y) * 2ll;
        res -= a[x] + a[y];
        cout << res << '\n';
      }
    }
  }
  int main() {
    int tc = 1;
    for (int t = 0; t < tc; t++) solve(t);
  }
  