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 i64 = long long;
  using u64 = unsigned long long;
  struct debug {
    ~debug() { cerr << endl; }
    template <class c>
    debug& operator<<(c x) {
      cerr << x;
      return *this;
    }
    template <class c, class d>
    debug& operator<<(pair<c, d> x) {
      *this << "(" << x.first << ", " << x.second << ")";
      return *this;
    }
    template <class c>
    debug& operator<<(vector<c> x) {
      *this << "[";
      int f = 0;
      for (auto it : x) {
        *this << (f ? ", " : "");
        *this << it;
        f = 1;
      }
      *this << "]";
      return *this;
    };
  };
  struct SegTree {
    vector<i64> tree;
    i64 n;
    SegTree(i64 _n) {
      n = _n;
      tree = vector<i64>(4 * n + 4);
    }
    i64 merge(i64 a, i64 b) { return a + b; }
    void update(i64 uIdx, i64 val, i64 idx, i64 l, i64 r) {
      if (l == r) {
        tree[idx] = val;
        return;
      }
      i64 mid = (l + r) / 2;
      if (uIdx <= mid)
        update(uIdx, val, 2 * idx, l, mid);
      else
        update(uIdx, val, 2 * idx + 1, mid + 1, r);
      tree[idx] = merge(tree[2 * idx], tree[2 * idx + 1]);
    }
    void update(i64 pos, i64 val) { update(pos, val, 1, 1, n); }
    i64 query(i64 ql, i64 qr, i64 idx, i64 l, i64 r) {
      if (ql > qr) return 0;
      if (ql == l && qr == r) {
        return tree[idx];
      }
      i64 mid = (l + r) / 2;
      return merge(query(ql, min(qr, mid), 2 * idx, l, mid),
                   query(max(ql, mid + 1), qr, 2 * idx + 1, mid + 1, r));
    }
    i64 query(i64 ql, i64 qr) { return query(ql, qr, 1, 1, n); }
  };
  const int N = 1e5 + 5;
  vector<int> G[N];
  vector<int> start(N);
  vector<int> finish(N);
  int timer = 1;
  int lca[N][24];
  void dfs(int cur, int p = 0) {
    lca[cur][0] = p;
    for (int i = 1; i < 22; i++) {
      lca[cur][i] = lca[lca[cur][i - 1]][i - 1];
    }
    start[cur] = timer++;
    for (int to : G[cur]) {
      if (to != p) {
        dfs(to, cur);
      }
    }
    finish[cur] = timer++;
  }
  bool isAncestor(int u, int v) {
    return start[u] <= start[v] && finish[u] >= finish[v];
  }
  int findLca(int u, int v) {
    if (isAncestor(u, v)) return u;
    if (isAncestor(v, u)) return v;
    for (int i = 21; i >= 0; i--) {
      if (!isAncestor(lca[u][i], v)) {
        u = lca[u][i];
      }
    }
    return lca[u][0];
  }
  int main() {
    finish[0] = 1000000000;
    int n, q;
    scanf("%d%d", &n, &q);
    vector<int> a(n + 3);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      a[i] = abs(a[i]);
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    dfs(1);
    SegTree tree(2 * n + 2);
    for (int i = 1; i <= n; i++) {
      tree.update(start[i], a[i]);
      tree.update(finish[i], -a[i]);
    }
    while (q--) {
      int type;
      scanf("%d", &type);
      if (type == 1) {
        int u, c;
        scanf("%d%d", &u, &c);
        c = abs(c);
        a[u] = c;
        tree.update(start[u], c);
        tree.update(finish[u], -c);
      } else {
        int u, v;
        scanf("%d%d", &u, &v);
        if (u == v) {
          puts("0");
        } else if (isAncestor(u, v)) {
          printf("%lld\n",
                 a[u] + a[v] + 2 * tree.query(start[u] + 1, start[v] - 1));
        } else if (isAncestor(v, u)) {
          printf("%lld\n",
                 a[u] + a[v] + 2 * tree.query(start[v] + 1, start[u] - 1));
        } else {
          int l = findLca(u, v);
          i64 ans = a[u] + a[l] + 2 * tree.query(start[l] + 1, start[u] - 1);
          ans += a[l] + a[v] + 2 * tree.query(start[l] + 1, start[v] - 1);
          printf("%lld\n", ans);
        }
      }
    }
  }
  