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 + 10;
  int a[maxn];
  int id2[maxn];
  struct seg_tree {
    long long sum[maxn << 2], add[maxn << 2];
    int ls(int x) { return x << 1; }
    int rs(int x) { return x << 1 | 1; }
    void push_up(int p) { sum[p] = sum[ls(p)] + sum[rs(p)]; }
    void build(int id, int l, int r) {
      add[id] = 0;
      if (l == r) {
        sum[id] = a[id2[l]];
        return;
      }
      int mid = (l + r) >> 1;
      build(ls(id), l, mid);
      build(rs(id), mid + 1, r);
      push_up(id);
    }
    void f(int id, int l, int r, long long k) {
      add[id] = add[id] + k;
      sum[id] = sum[id] + k * (r - l + 1);
    }
    void push_down(int id, int l, int r) {
      int mid = (l + r) >> 1;
      f(ls(id), l, mid, add[id]);
      f(rs(id), mid + 1, r, add[id]);
      add[id] = 0;
    }
    void iadd(int id, int l, int r, int x, int y, long long k) {
      if (x > y || x > r || y < l) return;
      if (x <= l && r <= y) {
        sum[id] += k * (r - l + 1);
        add[id] += k;
        return;
      }
      push_down(id, l, r);
      int mid = (l + r) >> 1;
      if (x <= mid) iadd(ls(id), l, mid, x, y, k);
      if (y > mid) iadd(rs(id), mid + 1, r, x, y, k);
      push_up(id);
    }
    long long iquery(int id, int l, int r, int x, int y) {
      if (x > y || x > r || y < l) return 0;
      long long ans = 0;
      if (x <= l && r <= y) return sum[id];
      int mid = (l + r) >> 1;
      push_down(id, l, r);
      if (x <= mid) ans += iquery(ls(id), l, mid, x, y);
      if (y > mid) ans += iquery(rs(id), mid + 1, r, x, y);
      return ans;
    }
    void upd(int id, int l, int r, int x, int w) {
      if (l == r) {
        sum[id] = w;
        return;
      }
      push_down(id, l, r);
      int mid = (l + r) / 2;
      if (x <= mid)
        upd(ls(id), l, mid, x, w);
      else
        upd(rs(id), mid + 1, r, x, w);
      push_up(id);
    }
  } seg;
  int n;
  vector<int> g[maxn];
  struct Tree_chain_subdivision {
    int cnt = 0;
    int w[maxn], wt[maxn], son[maxn], dfn[maxn], fa[maxn];
    int dep[maxn], siz[maxn], top[maxn];
    void init() {
      cnt = 0;
      dfs1(1, 1, 1);
      dfs2(1, 1);
      seg.build(1, 1, n);
    }
    void dfs1(int x, int f, int deep) {
      dep[x] = deep;
      fa[x] = f;
      siz[x] = 1;
      son[x] = 0;
      int maxson = -1;
      for (auto y : g[x]) {
        if (y == f) continue;
        dfs1(y, x, deep + 1);
        siz[x] += siz[y];
        if (siz[y] > maxson) son[x] = y, maxson = siz[y];
      }
    }
    void dfs2(int x, int topf) {
      dfn[x] = ++cnt;
      id2[cnt] = x;
      wt[cnt] = w[x];
      top[x] = topf;
      if (!son[x]) return;
      dfs2(son[x], topf);
      for (auto y : g[x]) {
        if (y == fa[x] || y == son[x]) continue;
        dfs2(y, y);
      }
    }
    int lca(int x, int y) {
      while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        x = fa[top[x]];
      }
      return dep[x] < dep[y] ? x : y;
    }
    void upd_range(int x, int y, int k) {
      while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        seg.iadd(1, 1, n, dfn[top[x]], dfn[x], k);
        x = fa[top[x]];
      }
      if (dep[x] > dep[y]) swap(x, y);
      seg.iadd(1, 1, n, dfn[x], dfn[y], k);
    }
    void upd(int u) { seg.upd(1, 1, n, dfn[u], a[u]); }
    long long q_range(int x, int y) {
      long long ans = 0;
      while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ans += seg.iquery(1, 1, n, dfn[top[x]], dfn[x]);
        x = fa[top[x]];
      }
      if (dep[x] > dep[y]) swap(x, y);
      ans += seg.iquery(1, 1, n, dfn[x], dfn[y]);
      return ans;
    }
    void upd_son(int x, int k) {
      seg.iadd(1, 1, n, dfn[x], dfn[x] + siz[x] - 1, k);
    }
    int q_son(int x) { return seg.iquery(1, 1, n, dfn[x], dfn[x] + siz[x] - 1); }
  } sp;
  int main() {
    std::ios::sync_with_stdio(false), cin.tie(0);
    int q;
    cin >> n >> q;
    for (int i = 1; i <= n; i++) cin >> a[i], a[i] = abs(a[i]);
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    sp.init();
    while (q--) {
      int op;
      cin >> op;
      if (op == 1) {
        int u, c;
        cin >> u >> c;
        c = abs(c);
        a[u] = c;
        sp.upd(u);
      } else {
        int u, v;
        cin >> u >> v;
        cout << sp.q_range(u, v) * 2 - a[u] - a[v] << '\n';
      }
    }
    return 0;
  }
  