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;
  using ll = long long;
  int dep[maxn + 10], sz[maxn + 10];
  int hson[maxn + 10], fa[maxn + 10];
  int top[maxn + 10];
  vector<int> edges[maxn + 10];
  int n, q;
  int dfsn[maxn + 10];
  int rdfsn[maxn + 10];
  int cnt = 0;
  ll a[maxn + 10];
  ll a2[maxn + 10];
  ll t[maxn + 10];
  void dfs1(int p, int fath) {
    int size = 1, mx = 0;
    dep[p] = dep[fath] + 1;
    fa[p] = fath;
    for (auto q : edges[p]) {
      if (q == fath) continue;
      a[q] = max(abs(t[p] + t[q]), abs(t[q] - t[p]));
      dfs1(q, p);
      size += sz[q];
      if (sz[q] > mx) {
        mx = sz[q];
        hson[p] = q;
      }
    }
    sz[p] = size;
  }
  void dfs2(int p) {
    dfsn[p] = ++cnt;
    rdfsn[cnt] = p;
    a2[cnt] = a[p];
    if (hson[p] != 0) {
      top[hson[p]] = top[p];
      dfs2(hson[p]);
    }
    for (auto q : edges[p]) {
      if (!top[q]) {
        top[q] = q;
        dfs2(q);
      }
    }
  }
  struct Segtree {
    ll val;
  } tree[(maxn << 2) + 10];
  void pushup(int rt) {
    tree[rt].val = tree[rt << 1].val + tree[rt << 1 | 1].val;
  }
  void build(int l = 1, int r = n, int rt = 1) {
    if (l == r) {
      tree[rt].val = a2[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  ll query(int L, int R, int l = 1, int r = n, int rt = 1) {
    if (L > r || R < l) return 0;
    if (L <= l && r <= R) {
      return tree[rt].val;
    }
    int mid = (l + r) >> 1;
    return query(L, R, l, mid, rt << 1) + query(L, R, mid + 1, r, rt << 1 | 1);
  }
  ll query_path(int x, int y) {
    ll ans = 0;
    while (top[x] != top[y]) {
      if (dep[top[x]] < dep[top[y]]) swap(x, y);
      ans = ans + query(dfsn[top[x]], dfsn[x]);
      x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ans = ans + query(dfsn[x] + 1, dfsn[y]);
    return ans;
  }
  void update(int L, int l = 1, int r = n, int rt = 1) {
    if (L == l && L == r) {
      int cur = rdfsn[L];
      int fath = fa[cur];
      tree[rt].val = max(abs(t[cur] + t[fath]), abs(t[cur] - t[fath]));
      return;
    }
    if (L < l || L > r) {
      return;
    }
    int mid = (l + r) / 2;
    if (L <= mid)
      update(L, l, mid, rt << 1);
    else if (L > mid)
      update(L, mid + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  void solve() {
    cin >> n >> q;
    for (int i = 1; i <= n; i++) {
      edges[i].clear();
      cnt = 0;
    }
    for (int i = 1; i <= n; i++) {
      cin >> t[i];
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      edges[u].emplace_back(v);
      edges[v].emplace_back(u);
    }
    dfs1(1, 0);
    top[1] = 1;
    dfs2(1);
    build();
    for (int i = 1; i <= q; i++) {
      int op = 0;
      cin >> op;
      if (op == 1) {
        int u, c;
        cin >> u >> c;
        t[u] = c;
        update(dfsn[u]);
        for (auto q : edges[u]) {
          if (q != fa[u]) update(dfsn[q]);
        }
      } else {
        int u, v;
        cin >> u >> v;
        cout << query_path(u, v) << "\n";
      }
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  