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;
  vector<long long> etour;
  vector<long long> adj[200005];
  long long preorder[200005];
  long long postorder[200005];
  long long timer = 0;
  long long dp[200005][20];
  long long a[200005];
  void DFS(long long u, long long par) {
    preorder[u] = timer;
    etour.push_back(a[u]);
    timer++;
    dp[u][0] = par;
    for (long long i = 1; i < 20; i++) {
      long long z = dp[u][i - 1];
      if (z == -1) {
        dp[u][i] = -1;
        continue;
      }
      dp[u][i] = dp[z][i - 1];
    }
    for (auto x : adj[u]) {
      if (x == par) {
        continue;
      }
      DFS(x, u);
    }
    postorder[u] = timer;
    etour.push_back(-a[u]);
    timer++;
  }
  long long is_ancestor(long long u, long long v) {
    if (preorder[u] <= preorder[v] && postorder[u] >= postorder[v]) {
      return 1;
    }
    return 0;
  }
  long long lca(long long u, long long v) {
    if (is_ancestor(u, v)) {
      return u;
    }
    long long z = u;
    for (long long i = 19; i >= 0; i--) {
      if (dp[z][i] == -1) {
        continue;
      }
      if (is_ancestor(dp[z][i], v)) {
        continue;
      }
      z = dp[z][i];
    }
    return dp[z][0];
  }
  struct segtree {
    vector<long long> v;
    void init(long long n) {
      long long curr = 1;
      while (curr < n) {
        curr *= 2;
      }
      for (long long i = 0; i < 2 * curr; i++) {
        v.push_back(0);
      }
    }
    void update(long long lx, long long rx, long long i, long long z,
                long long curr) {
      if (rx - lx == 1) {
        v[curr] = z;
        return;
      }
      long long m = (lx + rx) / 2;
      if (i < m) {
        update(lx, m, i, z, 2 * curr + 1);
      } else {
        update(m, rx, i, z, 2 * curr + 2);
      }
      v[curr] = v[2 * curr + 1] + v[2 * curr + 2];
    }
    long long getsum(long long lx, long long rx, long long l, long long r,
                     long long curr) {
      if (lx >= l && rx <= r) {
        return v[curr];
      }
      if (lx >= r || rx <= l) {
        return 0;
      }
      long long m = (lx + rx) / 2;
      return getsum(lx, m, l, r, 2 * curr + 1) +
             getsum(m, rx, l, r, 2 * curr + 2);
    }
  };
  long long bforce(long long u, long long v) {
    queue<long long> q;
    q.push(u);
    long long dist[100005];
    long long visited[100005];
    memset(visited, 0, sizeof(visited));
    visited[u] = 1;
    dist[u] = a[u];
    while (!q.empty()) {
      long long z = q.front();
      for (auto x : adj[z]) {
        if (visited[x]) {
          continue;
        }
        dist[x] = dist[z] + a[x];
        q.push(x);
        visited[x] = 1;
      }
      q.pop();
    }
    return 2 * dist[v] - (a[u] + a[v]);
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long n, q;
    cin >> n >> q;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      u--;
      v--;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    DFS(0, -1);
    struct segtree st;
    long long z = etour.size();
    assert(z == (2 * n));
    st.init(z);
    for (long long i = 0; i < z; i++) {
      st.update(0, z, i, etour[i], 0);
    }
    while (q--) {
      long long qtype, u, v;
      cin >> qtype >> u >> v;
      if (qtype == 1) {
        v = abs(v);
        u--;
        a[u] = v;
        st.update(0, z, preorder[u], v, 0);
        st.update(0, z, postorder[u], -v, 0);
      } else {
        u--;
        v--;
        long long zz = lca(u, v);
        long long pok = st.getsum(0, z, preorder[zz], preorder[u] + 1, 0);
        long long yo = st.getsum(0, z, preorder[zz], preorder[v] + 1, 0);
        pok += yo;
        pok -= a[zz];
        pok = pok * 2;
        pok -= a[u];
        pok -= a[v];
        cout << pok << "\n";
      }
    }
  }
  