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 N = 2e5 + 5, L = 23;
  vector<int> g[N];
  int a[N], n, q;
  int timer;
  vector<int> tin(N), tout(N);
  vector<vector<int>> up(N, vector<int>(L));
  int par[N], id[N], _id[N], cid = 0, hvc[N], head[N], sz[N], depth[N];
  struct SegmentTree {
    struct node {
      long long val = 0, lz = 0;
      void apply(int l, int r, int d) {
        val = d;
        lz = d;
      }
    };
    node merge(const node &a, const node &b) {
      node res;
      res.val = a.val + b.val;
      return res;
    }
    inline void push(int v, int l, int r) {
      int m = (l + r) / 2, u = (m - l + 1) * 2 + v;
      if (tr[v].lz != 0) {
        tr[v + 1].apply(l, m, tr[v].lz);
        tr[u].apply(m + 1, r, tr[v].lz);
        tr[v].lz = 0;
      }
    }
    int n;
    vector<node> tr;
    void build(int v, int l, int r) {
      if (l == r) {
        return;
      }
      int m = (l + r) / 2, u = (m - l + 1) * 2 + v;
      build(v + 1, l, m), build(u, m + 1, r);
      tr[v] = merge(tr[v + 1], tr[u]);
    }
    void build(int v, int l, int r, vector<int> &d) {
      if (l == r) {
        tr[v].apply(l, r, d[l]);
        return;
      }
      int m = (l + r) / 2, u = (m - l + 1) * 2 + v;
      build(v + 1, l, m, d), build(u, m + 1, r, d);
      tr[v] = merge(tr[v + 1], tr[u]);
    }
    void udt(int v, int l, int r, int L, int R, int d) {
      if (L <= l and r <= R) {
        tr[v].apply(l, r, d);
        return;
      }
      int m = (l + r) / 2, u = (m - l + 1) * 2 + v;
      push(v, l, r);
      if (L <= m) udt(v + 1, l, m, L, R, d);
      if (R > m) udt(u, m + 1, r, L, R, d);
      tr[v] = merge(tr[v + 1], tr[u]);
    }
    node get(int v, int l, int r, int L, int R) {
      if (L <= l and r <= R) return tr[v];
      int m = (l + r) / 2, u = (m - l + 1) * 2 + v;
      push(v, l, r);
      if (R <= m)
        return get(v + 1, l, m, L, R);
      else if (L > m)
        return get(u, m + 1, r, L, R);
      return merge(get(v + 1, l, m, L, R), get(u, m + 1, r, L, R));
    }
    SegmentTree(int _n) : n(_n) {
      assert(n > 0);
      tr.resize(2 * n);
      build(0, 0, n - 1);
    }
    SegmentTree(vector<int> &d) {
      n = d.size();
      assert(n > 0);
      tr.resize(2 * n);
      build(0, 0, n - 1, d);
    }
    long long get(int l, int r) {
      assert(0 <= l and l <= r and r <= n - 1);
      return get(0, 0, n - 1, l, r).val;
    }
    void udt(int l, int r, int d) {
      assert(0 <= l and l <= r and r <= n - 1);
      return udt(0, 0, n - 1, l, r, d);
    }
  };
  SegmentTree st(N);
  void dfs(int v, int p) {
    par[v] = p, sz[v] = 1;
    tin[v] = ++timer;
    up[v][0] = p;
    for (int i = 1; i < L; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
    int mx = 0;
    for (auto u : g[v]) {
      if (u == p) continue;
      dfs(u, v);
      depth[u] = depth[v] + 1;
      sz[v] += sz[u];
      if (sz[u] > mx) {
        mx = sz[u];
        hvc[v] = u;
      }
    }
    tout[v] = ++timer;
  }
  void decompose(int v) {
    id[v] = cid;
    _id[cid++] = v;
    if (hvc[v] != -1) {
      head[hvc[v]] = head[v];
      decompose(hvc[v]);
    }
    for (auto u : g[v]) {
      if (u == par[v] or u == hvc[v]) continue;
      head[u] = u;
      decompose(u);
    }
  }
  bool is_ancestor(int u, int v) {
    return tin[u] <= tin[v] && tout[u] >= tout[v];
  }
  int lca(int u, int v) {
    if (is_ancestor(u, v)) return u;
    if (is_ancestor(v, u)) return v;
    for (int i = L - 1; i >= 0; --i) {
      if (!is_ancestor(up[u][i], v)) u = up[u][i];
    }
    return up[u][0];
  }
  void update(int v, int x) { st.udt(id[v], id[v], x); }
  long long query(int u, int v) {
    long long res = 0;
    while (head[u] != head[v]) {
      if (depth[head[u]] > depth[head[v]]) swap(u, v);
      res += st.get(id[head[v]], id[v]);
      v = par[head[v]];
    }
    if (depth[u] > depth[v]) swap(u, v);
    res += st.get(id[u], id[v]);
    return res;
  }
  int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    memset(hvc, -1, sizeof hvc);
    cin >> n >> q;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i] = abs(a[i]);
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      --u, --v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(0, 0);
    decompose(0);
    for (int i = 0; i < n; i++) {
      st.udt(id[i], id[i], a[i]);
    }
    while (q--) {
      int t;
      cin >> t;
      if (t == 1) {
        int v, x;
        cin >> v >> x;
        --v;
        x = abs(x);
        a[v] = x;
        update(v, x);
      } else {
        int u, v;
        cin >> u >> v;
        --u, --v;
        int l = lca(u, v);
        cout << 2 * (query(0, u) + query(0, v) - 2 * query(0, l) + a[l]) - a[u] -
                    a[v]
             << '\n';
      }
    }
  }
  