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 int MAXN = 1e5 + 5;
  vector<int> adj[MAXN];
  int sz[MAXN], dep[MAXN], top[MAXN], son[MAXN], rk[MAXN], dfn[MAXN], dfs_clock,
      fa[MAXN], n, q;
  long long c[MAXN], sum[MAXN << 2];
  void build(int rt, int l, int r) {
    if (l == r) {
      sum[rt] = c[rk[l]];
      return;
    }
    int mid = (l + r) >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
  }
  void upd(int rt, int l, int r, int pos, long long x) {
    if (l == r) {
      sum[rt] = x;
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) {
      upd(rt << 1, l, mid, pos, x);
    } else {
      upd(rt << 1 | 1, mid + 1, r, pos, x);
    }
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
  }
  long long qry(int rt, int l, int r, int L, int R) {
    if (L <= l && R >= r) {
      return sum[rt];
    }
    int mid = (l + r) >> 1;
    if (R <= mid) {
      return qry(rt << 1, l, mid, L, R);
    } else if (L > mid) {
      return qry(rt << 1 | 1, mid + 1, r, L, R);
    } else {
      return qry(rt << 1, l, mid, L, mid) +
             qry(rt << 1 | 1, mid + 1, r, mid + 1, R);
    }
  }
  void dfs1(int u, int d) {
    sz[u] = 1;
    son[u] = 0;
    dep[u] = d;
    for (int v : adj[u])
      if (fa[u] != v) {
        fa[v] = u;
        dfs1(v, d + 1);
        sz[u] += sz[v];
        if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;
      }
  }
  void dfs2(int u, int t) {
    top[u] = t;
    dfn[u] = ++dfs_clock;
    rk[dfs_clock] = u;
    if (son[u]) dfs2(son[u], t);
    for (int v : adj[u])
      if (son[u] != v && fa[u] != v) {
        dfs2(v, v);
      }
  }
  long long get(int u, int v) {
    long long res = 0;
    while (top[u] != top[v]) {
      if (dep[top[u]] < dep[top[v]]) {
        swap(u, v);
      }
      res += qry(1, 1, n, dfn[top[u]], dfn[u]);
      u = fa[top[u]];
    }
    if (dfn[u] > dfn[v]) swap(u, v);
    res += qry(1, 1, n, dfn[u], dfn[v]);
    return res;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> q;
    for (int i = 1; i <= n; i++) {
      cin >> c[i];
      c[i] = abs(c[i]);
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    dfs1(1, 0);
    dfs2(1, 1);
    build(1, 1, n);
    for (int i = 1; i <= q; i++) {
      int op;
      cin >> op;
      if (op == 1) {
        int u;
        int x;
        cin >> u >> x;
        x = abs(x);
        upd(1, 1, n, dfn[u], 1ll * x);
        c[u] = x;
      } else {
        int u, v;
        cin >> u >> v;
        cout << 2ll * get(u, v) - c[u] - c[v] << '\n';
      }
    }
    return 0;
  }
  