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;
  struct segtree {
    vector<long long> v;
    long long size;
    void init(long long n) {
      size = 1;
      while (size < n) size *= 2;
      v.resize(size * 2, 0);
    }
    void set(long long x, long long val) { set(x, val, 0, 0, size); }
    void set(long long x, long long val, long long cur, long long lx,
             long long rx) {
      if (rx - lx == 1) {
        v[cur] = val;
      } else {
        long long mid = (lx + rx) / 2;
        if (x < mid) {
          set(x, val, 2 * cur + 1, lx, mid);
        } else {
          set(x, val, 2 * cur + 2, mid, rx);
        }
        v[cur] = v[2 * cur + 1] + v[2 * cur + 2];
      }
    }
    long long get(long long l, long long r) { return get(l, r, 0, 0, size); }
    long long get(long long l, long long r, long long cur, long long lx,
                  long long rx) {
      if (l >= rx or r <= lx) {
        return (long long)0;
      }
      if (l <= lx and r >= rx) {
        return v[cur];
      }
      long long m = (lx + rx) / 2;
      return (get(l, r, 2 * cur + 1, lx, m) + get(l, r, 2 * cur + 2, m, rx));
    }
  };
  struct HLD {
    long long cnt;
    vector<long long> depth, sz, parent, tp, idx;
    segtree st;
    HLD(vector<vector<long long> > &gr, long long root) {
      long long n = gr.size();
      depth.resize(n, -1);
      st.init(n + 1);
      sz.resize(n, 0);
      parent.resize(n, 0);
      tp.resize(n, 0);
      idx.resize(n, 0);
      dfs_sz(gr, root, root);
      cnt = 0;
      dfs_hld(gr, root, root, root);
    }
    long long dfs_sz(vector<vector<long long> > &gr, long long cur,
                     long long par) {
      depth[cur] = depth[par] + 1;
      sz[cur] = 1;
      parent[cur] = par;
      for (auto x : gr[cur]) {
        if (x == par) continue;
        sz[cur] += dfs_sz(gr, x, cur);
      }
      return sz[cur];
    }
    void dfs_hld(vector<vector<long long> > &gr, long long cur, long long par,
                 long long top) {
      tp[cur] = top;
      idx[cur] = cnt++;
      long long mx_sz = -1, mx_child = -1;
      for (auto x : gr[cur]) {
        if (x == par) continue;
        if (sz[x] > mx_sz) {
          mx_sz = sz[x];
          mx_child = x;
        }
      }
      if (mx_child == -1) return;
      dfs_hld(gr, mx_child, cur, top);
      for (auto x : gr[cur]) {
        if (x == par or x == mx_child) continue;
        dfs_hld(gr, x, cur, x);
      }
    }
    long long path(long long x, long long y) {
      long long ans =
          -((st.get(idx[x], idx[x] + 1) + st.get(idx[y], idx[y] + 1)) / 2ll);
      while (tp[x] != tp[y]) {
        if (depth[tp[x]] < depth[tp[y]]) swap(x, y);
        ans += st.get(idx[tp[x]], idx[x] + 1);
        x = parent[tp[x]];
      }
      if (depth[x] > depth[y]) swap(x, y);
      ans += st.get(idx[x], idx[y] + 1);
      return ans;
    }
  };
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long n, q;
    cin >> n >> q;
    vector<vector<long long> > gr(n);
    vector<long long> a(n);
    for (auto &x : a) cin >> x;
    for (long long i = 0; i + 1 < n; ++i) {
      long long x, y;
      cin >> x >> y;
      --x;
      --y;
      gr[x].push_back(y);
      gr[y].push_back(x);
    }
    HLD h(gr, 0);
    for (long long i = 0; i < n; ++i) {
      h.st.set(h.idx[i], 2ll * abs(a[i]));
    }
    while (q--) {
      long long ch, x, y;
      cin >> ch >> x >> y;
      if (ch == 1) {
        h.st.set(h.idx[x - 1], 2ll * abs(y));
      } else {
        --x;
        --y;
        cout << h.path(x, y) << "\n";
      }
    }
    return 0;
  }
  