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;
  mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  mt19937 rnf(2106);
  const int N = 100005;
  int n, qq;
  int a[N];
  vector<int> g[N];
  int q[N];
  int p0[N];
  void dfs0(int x, int p) {
    p0[x] = p;
    q[x] = 1;
    for (int i = 0; i < g[x].size(); ++i) {
      int h = g[x][i];
      if (h == p) continue;
      dfs0(h, x);
      q[x] += q[h];
    }
    for (int i = 0; i < g[x].size(); ++i) {
      int h = g[x][i];
      if (h == p) swap(g[x][i], g[x].back());
    }
    for (int i = 0; i < g[x].size(); ++i) {
      int h = g[x][i];
      if (h == p) continue;
      if (q[h] > q[g[x][0]]) swap(g[x][i], g[x][0]);
    }
  }
  int f[N];
  int tin[N], tout[N], ti;
  void dfs1(int x, int p) {
    tin[x] = ++ti;
    for (int i = 0; i < g[x].size(); ++i) {
      int h = g[x][i];
      if (h == p) continue;
      if (i == 0) {
        f[h] = f[x];
        dfs1(h, x);
      } else {
        f[h] = h;
        dfs1(h, x);
      }
    }
    tout[x] = ti;
  }
  long long t[N * 4];
  void ubd(int tl, int tr, int x, int y, int pos) {
    if (tl == tr) {
      t[pos] = y;
      return;
    }
    int m = (tl + tr) / 2;
    if (x <= m)
      ubd(tl, m, x, y, pos * 2);
    else
      ubd(m + 1, tr, x, y, pos * 2 + 1);
    t[pos] = t[pos * 2] + t[pos * 2 + 1];
  }
  long long qry(int tl, int tr, int l, int r, int pos) {
    if (l > r) return 0;
    if (tl == l && tr == r) return t[pos];
    int m = (tl + tr) / 2;
    return (qry(tl, m, l, min(m, r), pos * 2) +
            qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
  }
  long long qryy(int x, int y) {
    long long ans = 0;
    while (1) {
      if (tin[x] < tin[y]) swap(x, y);
      if (f[x] == f[y]) {
        ans += qry(1, n, tin[y], tin[x], 1);
        return ans;
      }
      ans += qry(1, n, tin[f[x]], tin[x], 1);
      x = p0[f[x]];
    }
  }
  void solv() {
    cin >> n >> qq;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (int i = 0; i < n - 1; ++i) {
      int x, y;
      cin >> x >> y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    dfs0(1, 1);
    f[1] = 1;
    dfs1(1, 1);
    for (int x = 1; x <= n; ++x) ubd(1, n, tin[x], a[x], 1);
    while (qq--) {
      int ty;
      cin >> ty;
      if (ty == 1) {
        int x, y;
        cin >> x >> y;
        a[x] = y;
        a[x] = abs(a[x]);
        ubd(1, n, tin[x], a[x], 1);
      } else {
        int x, y;
        cin >> x >> y;
        cout << qryy(x, y) * 2 - a[x] - a[y] << "\n";
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int tt = 1;
    while (tt--) solv();
    return 0;
  }
  