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 long long linf = 4000000000000000000LL;
  const long long inf = 1000000007;
  void pv(vector<int> a) {
    for (auto& x : a) cout << x << " ";
    cout << '\n';
  }
  void pv(vector<long long> a) {
    for (auto& x : a) cout << x << " ";
    cout << '\n';
  }
  void pv(vector<vector<int>> a) {
    for (int i = (0); i < (int(a.size())); ++i) {
      cout << i << '\n';
      pv(a[i]);
      cout << '\n';
    }
  }
  void pv(vector<vector<long long>> a) {
    for (int i = (0); i < (int(a.size())); ++i) {
      cout << i << '\n';
      pv(a[i]);
    }
    cout << '\n';
  }
  void pv(vector<string> a) {
    for (auto& x : a) cout << x << '\n';
    cout << '\n';
  }
  void setIO(string second) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  }
  template <int SZ>
  struct hld {
    struct item {
      long long sum;
    };
    item single(long long i) { return {i}; }
    item merge(item x, item y) {
      item ans;
      ans.sum = x.sum + y.sum;
      return ans;
    }
    vector<item> tree;
    vector<item> A;
    int height;
    item neutral = {0};
    void build(vector<item>& A, int v, int tl, int tr) {
      if (tl == tr)
        tree[v] = A[tl];
      else {
        int mid = (tl + tr) / 2;
        build(A, 2 * v + 1, tl, mid);
        build(A, 2 * v + 2, mid + 1, tr);
        tree[v] = merge(tree[2 * v + 1], tree[2 * v + 2]);
      }
    }
    void build(vector<long long>& B) {
      int n = B.size();
      height = log2(n + 1) + 1;
      A.resize(n);
      tree.resize((1 << height + 1) - 1);
      for (int i = (0); i < (n); ++i) A[i] = single(B[i]);
      A.resize(1 << height, neutral);
      build(A, 0, 0, (1 << height) - 1);
    }
    item query(int v, int tl, int tr, int l, int r) {
      if (r < tl || l > tr) return neutral;
      if (l <= tl && r >= tr) {
        return tree[v];
      }
      int mid = (tl + tr) / 2;
      return merge(query(2 * v + 1, tl, mid, l, r),
                   query(2 * v + 2, mid + 1, tr, l, r));
    }
    long long query(int l, int r) {
      return query(0, 0, (1 << height) - 1, l, r).sum;
    }
    void update(int v, int tl, int tr, int pos, item val) {
      if (tl == tr) {
        tree[v] = val;
      } else {
        int mid = (tl + tr) / 2;
        if (pos <= mid)
          update(2 * v + 1, tl, mid, pos, val);
        else
          update(2 * v + 2, mid + 1, tr, pos, val);
        tree[v] = merge(tree[2 * v + 1], tree[2 * v + 2]);
      }
    }
    void update(int pos, long long val) {
      update(0, 0, (1 << height) - 1, pos, single(val));
    }
    int n;
    vector<vector<int>> adj;
    vector<long long> a;
    vector<int> label, sizes, depth, heavy_child, parent, head, dist;
    vector<vector<int>> jump;
    int get(int a, int b) {
      for (int j = (0); j < (SZ); ++j) {
        if (b & (1 << j)) {
          a = jump[a][j];
        }
      }
      return a;
    }
    int LCA(int a, int b) {
      if (depth[a] < depth[b]) swap(a, b);
      a = get(a, depth[a] - depth[b]);
      if (a == b) return a;
      int j = SZ;
      while (j) {
        j--;
        int ta = jump[a][j], tb = jump[b][j];
        if (ta != tb) a = ta, b = tb;
      }
      return parent[a];
    }
    void dfs0(int i, int p) {
      parent[i] = p;
      sizes[i] = 1;
      for (auto& x : adj[i]) {
        if (x == p) continue;
        depth[x] = depth[i] + 1;
        dfs0(x, i);
        sizes[i] += sizes[x];
      }
    }
    int cur = 0;
    void dfs1(int i, int p) {
      label[i] = cur++;
      if (heavy_child[i] != -1) {
        head[heavy_child[i]] = head[i];
        dfs1(heavy_child[i], i);
        dist[i] = 1 + dist[heavy_child[i]];
      }
      for (auto& x : adj[i]) {
        if (x != heavy_child[i] && x != p) {
          head[x] = x;
          dfs1(x, i);
        }
      }
    }
    void init(vector<long long>& y, vector<vector<int>>& x) {
      adj = x;
      n = int(adj.size());
      a = y;
      sizes.resize(n);
      depth.resize(n);
      parent.resize(n);
      dfs0(0, -1);
      jump.resize(n, vector<int>(SZ));
      for (int i = (0); i < (n); ++i) jump[i][0] = parent[i];
      for (int i = (0); i < (SZ - 1); ++i) {
        for (int j = (0); j < (n); ++j)
          jump[j][i + 1] = (jump[j][i] == -1 ? -1 : jump[jump[j][i]][i]);
      }
      heavy_child.resize(n);
      for (int i = (0); i < (n); ++i) {
        vector<pair<int, int>> t;
        for (auto& x : adj[i])
          if (parent[i] != x) t.push_back({sizes[x], x});
        sort(t.begin(), t.end(), greater<pair<int, int>>());
        if (int(t.size()))
          heavy_child[i] = t[0].second;
        else
          heavy_child[i] = -1;
      }
      label.resize(n);
      head.resize(n);
      dist.resize(n);
      dfs1(0, -1);
      vector<long long> t(n);
      for (int i = (0); i < (n); ++i) t[label[i]] = y[i];
      build(t);
    }
    long long solve(int x, int l) {
      long long ans = 0;
      while (1) {
        if (x == l) return ans;
        if (head[x] == x) {
          ans += a[x];
          x = parent[x];
          continue;
        }
        int h = head[x];
        if (depth[h] > depth[l]) {
          ans += query(label[h], label[x]);
          x = parent[h];
          continue;
        }
        ans += query(label[l] + 1, label[x]);
        x = l;
      }
      return ans;
    }
    long long query_path(int x, int y) {
      int l = LCA(x, y);
      return solve(x, l) + solve(y, l) + a[l];
    }
    void update_path(int x, int y) {
      a[x] = y;
      update(label[x], y);
    }
  };
  int n, q;
  vector<long long> a;
  vector<vector<int>> adj;
  int main() {
    setIO("test1");
    cin >> n >> q;
    a.resize(n);
    for (auto& x : a) cin >> x, x = abs(x);
    adj.resize(n);
    vector<long long> t(n);
    for (int i = (0); i < (n); ++i) t[i] = 2 * a[i];
    for (int i = (0); i < (n - 1); ++i) {
      int x, y;
      cin >> x >> y;
      x--, y--;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    hld<20> g;
    g.init(t, adj);
    for (int i = (0); i < (q); ++i) {
      int t, x, y;
      cin >> t >> x >> y;
      if (t == 1) {
        y = abs(y);
        x--;
        a[x] = y;
        g.update_path(x, 2 * y);
      } else {
        x--, y--;
        cout << g.query_path(x, y) - a[x] - a[y] << '\n';
      }
    }
  }
  