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;
  long long n, m, r, p;
  long long fa[200005];
  long long dfn[200005];
  long long rk[200005];
  long long dep[200005];
  long long size1[200005];
  long long son[200005];
  long long top[200005];
  long long bottom[200005];
  long long cnt;
  vector<long long> v[200005];
  void dfs1(long long now, long long fanow) {
    size1[now] = 1;
    dep[now] = dep[fanow] + 1;
    long long max1 = 0;
    long long max2 = 0;
    for (long long i = 0; i < v[now].size(); i++) {
      if (v[now][i] != fanow) {
        fa[v[now][i]] = now;
        dfs1(v[now][i], now);
        size1[now] += size1[v[now][i]];
        if (size1[v[now][i]] > max1) {
          max1 = size1[v[now][i]];
          max2 = v[now][i];
        }
      }
    }
    son[now] = max2;
    return;
  }
  void dfs2(long long now, long long fanow) {
    dfn[now] = ++cnt;
    rk[cnt] = now;
    top[son[now]] = top[now];
    bottom[now] = dfn[now];
    if (son[now]) {
      dfs2(son[now], now);
      bottom[now] = max(bottom[now], bottom[son[now]]);
    }
    for (long long i = 0; i < v[now].size(); i++) {
      if (v[now][i] != fanow && v[now][i] != son[now]) {
        top[v[now][i]] = v[now][i];
        dfs2(v[now][i], now);
        bottom[now] = max(bottom[now], bottom[v[now][i]]);
      }
    }
    return;
  }
  struct node {
    long long sum;
    long long l, r;
    long long lazy;
  } node[4 * 200005];
  long long a[200005];
  void pushup(long long now) {
    node[now].sum = node[now * 2].sum + node[now * 2 + 1].sum;
    return;
  }
  void build(long long now, long long l, long long r) {
    node[now].l = l;
    node[now].r = r;
    if (l == r) {
      node[now].sum = a[rk[l]];
      return;
    }
    long long mid = (l + r) >> 1;
    build(now * 2, l, mid);
    build(now * 2 + 1, mid + 1, r);
    pushup(now);
    return;
  }
  void change(long long now, long long l, long long r, long long x, long long y,
              long long z) {
    if (l >= x && r <= y) {
      node[now].sum = z;
      return;
    }
    long long mid = (l + r) >> 1;
    if (x <= mid) change(now * 2, l, mid, x, y, z);
    if (y > mid) change(now * 2 + 1, mid + 1, r, x, y, z);
    pushup(now);
    return;
  }
  long long query(long long now, long long l, long long r, long long x,
                  long long y) {
    if (l >= x && r <= y) {
      return node[now].sum;
    }
    long long mid = (l + r) >> 1;
    long long ans = 0;
    if (x <= mid) ans += query(now * 2, l, mid, x, y);
    if (y > mid) ans += query(now * 2 + 1, mid + 1, r, x, y);
    return ans;
  }
  void change(long long l, long long r, long long z) {
    while (top[l] != top[r]) {
      if (dep[top[l]] < dep[top[r]]) {
        swap(l, r);
      }
      change(1, 1, n, dfn[top[l]], dfn[l], z);
      l = fa[top[l]];
    }
    if (dep[l] < dep[r]) {
      swap(l, r);
    }
    change(1, 1, n, dfn[r], dfn[l], z);
    return;
  }
  long long query(long long l, long long r) {
    long long ans = 0;
    while (top[l] != top[r]) {
      if (dep[top[l]] < dep[top[r]]) {
        swap(l, r);
      }
      ans += query(1, 1, n, dfn[top[l]], dfn[l]);
      l = fa[top[l]];
    }
    if (dep[l] < dep[r]) {
      swap(l, r);
    }
    ans += query(1, 1, n, dfn[r], dfn[l]);
    return ans;
  }
  void change(long long now, long long z) {
    change(1, 1, n, dfn[now], bottom[now], z);
    return;
  }
  long long query(long long now) { return query(1, 1, n, dfn[now], bottom[now]); }
  long long lca(long long l, long long r) {
    while (top[l] != top[r]) {
      if (dep[top[l]] < dep[top[r]]) {
        swap(l, r);
      }
      l = fa[top[l]];
    }
    if (dep[l] < dep[r]) return l;
    return r;
  }
  signed main() {
    cin >> n >> m;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      a[i] = llabs(a[i]);
    }
    for (long long i = 1; i <= n - 1; i++) {
      long long u, v1;
      cin >> u >> v1;
      v[u].push_back(v1);
      v[v1].push_back(u);
    }
    top[1] = 1;
    dfs1(1, 0);
    dfs2(1, 0);
    build(1, 1, n);
    for (long long i = 1; i <= m; i++) {
      long long opt;
      cin >> opt;
      if (opt == 1) {
        long long u, w;
        cin >> u >> w;
        w = llabs(w);
        a[u] = w;
        change(u, u, w);
      } else {
        long long u, vv;
        cin >> u >> vv;
        long long temp1 = 2 * query(u, vv) - a[u] - a[vv];
        cout << temp1 << endl;
      }
    }
    return 0;
  }
  