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;
  int n, q;
  const long long N = 2e5 + 100;
  vector<long long> v[N];
  long long dfn[N], num[N], val[N], sz[N], hs[N], tp[N], dep[N], f[N];
  inline void dfs(long long x, long long fa) {
    sz[x] = 1;
    long long num = 0;
    dep[x] = dep[fa] + 1;
    f[x] = fa;
    for (auto y : v[x]) {
      if (y == fa) continue;
      dfs(y, x);
      sz[x] += sz[y];
      if (num < sz[y]) hs[x] = y, num = sz[y];
    }
  }
  long long cnt;
  inline void redfs(long long x, long long fa, long long top) {
    tp[x] = top;
    dfn[x] = ++cnt;
    num[cnt] = x;
    if (hs[x]) redfs(hs[x], x, top);
    for (auto y : v[x]) {
      if (y == fa || y == hs[x]) continue;
      redfs(y, x, y);
    }
  }
  struct seg {
    long long l, r, v;
  } t[N << 2];
  long long a[N];
  inline void build(long long p, long long l, long long r) {
    t[p].l = l, t[p].r = r;
    if (l == r) {
      t[p].v = abs(val[num[l]]);
      return;
    }
    long long mid = l + r >> 1;
    build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r);
    t[p].v = t[p << 1].v + t[p << 1 | 1].v;
  }
  inline void modify(long long p, long long x, long long v) {
    if (t[p].l == t[p].r) {
      t[p].v = v;
      return;
    }
    long long mid = t[p].l + t[p].r >> 1;
    if (x <= mid)
      modify(p << 1, x, v);
    else
      modify(p << 1 | 1, x, v);
    t[p].v = t[p << 1].v + t[p << 1 | 1].v;
  }
  inline long long query(long long p, long long x, long long y) {
    if (x <= t[p].l && t[p].r <= y) {
      return t[p].v;
    }
    long long mid = t[p].l + t[p].r >> 1;
    if (y <= mid) return query(p << 1, x, y);
    if (x > mid) return query(p << 1 | 1, x, y);
    return query(p << 1, x, y) + query(p << 1 | 1, x, y);
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin >> n >> q;
    for (long long i = 1; i <= n; i++) cin >> val[i], val[i] = abs(val[i]);
    for (long long i = 1; i <= n - 1; i++) {
      long long x, y;
      cin >> x >> y;
      v[x].push_back(y), v[y].push_back(x);
    }
    dfs(1, 0);
    redfs(1, 0, 1);
    build(1, 1, n);
    for (long long i = 1; i <= q; i++) {
      long long opt, x, y;
      cin >> opt >> x >> y;
      if (opt == 1) {
        modify(1, dfn[x], abs(y));
        val[x] = abs(y);
      } else {
        long long ans = 0, X = x, Y = y;
        ;
        while (tp[x] != tp[y]) {
          if (dep[tp[x]] < dep[tp[y]]) swap(x, y);
          ans += query(1, dfn[tp[x]], dfn[x]);
          x = f[tp[x]];
        }
        if (dep[x] < dep[y]) swap(x, y);
        ans += query(1, dfn[y], dfn[x]);
        cout << ans * 2 - val[X] - val[Y] << "\n";
      }
    }
  }
  