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;
  inline void chmax(long long &x, long long y) { x = max(x, y); }
  inline void chmin(long long &x, long long y) { x = min(x, y); }
  inline void pls(long long &x, long long y, long long Mod) {
    x = (x + y >= Mod) ? (x + y - Mod) : (x + y);
  }
  inline void sub(long long &x, long long y, long long Mod) {
    x = (x - y < 0) ? (x - y + Mod) : (x - y);
  }
  long long max(long long a, long long b) { return a > b ? a : b; }
  long long min(long long a, long long b) { return a < b ? a : b; }
  long long lowbit(long long x) { return x & (-x); }
  long long prime(long long x) {
    if (x <= 1) return 0;
    for (int i = 2; i <= int(sqrt(x)); i++) {
      if (x % i == 0) return 0;
    }
    return 1;
  }
  bool cmp(long long a, long long b) { return a > b; }
  long long gcd(long long a, long long b) {
    long long r;
    while (b > 0) {
      r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  long long powmod(long long a, long long b, long long mod) {
    long long res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
    }
    return res;
  }
  inline int readint() {
    int f = 1, x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    return f * x;
  }
  inline long long readll() {
    long long f = 1, x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    return f * x;
  }
  void fl(string name) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
  const int dx[8] = {-1, 1, 0, 0, -1, 1, -1, 1},
            dy[8] = {0, 0, -1, 1, -1, -1, 1, 1};
  struct EDGE {
    int nxt, to;
  } edge[100010 << 1];
  int n, m, head[100010], tot, a[100010], in[100010];
  void add_edge(int u, int v) {
    edge[++tot].nxt = head[u];
    edge[tot].to = v;
    head[u] = tot;
  }
  int fa[100010], son[100010], dep[100010], t[100010];
  int top[100010], siz[100010], dfn[100010], tick;
  void dfs1(int u, int ff) {
    fa[u] = ff;
    dep[u] = dep[ff] + 1;
    siz[u] = 1;
    for (int i = head[u]; i; i = edge[i].nxt) {
      int v = edge[i].to;
      if (v == ff) continue;
      dfs1(v, u);
      siz[u] += siz[v];
      if (siz[son[u]] < siz[v]) {
        son[u] = v;
      }
    }
  }
  int calc(int u, int v) { return max(abs(u + v), abs(u - v)); }
  void dfs2(int u, int ff) {
    dfn[u] = ++tick;
    t[dfn[u]] = calc(a[u], a[fa[u]]);
    top[u] = ff;
    if (son[u]) dfs2(son[u], ff);
    for (int i = head[u]; i; i = edge[i].nxt) {
      int v = edge[i].to;
      if (v == fa[u] || v == son[u]) continue;
      dfs2(v, v);
    }
  }
  long long sum[100010];
  void add(int x, int k) {
    for (; x <= n; sum[x] += k, x += lowbit(x))
      ;
  }
  long long query(int x) {
    long long ans = 0;
    for (; x; ans += sum[x], x ^= lowbit(x))
      ;
    return ans;
  }
  void modify(int x, int k) {
    a[x] = k;
    int qwq = calc(a[x], a[fa[x]]);
    add(dfn[x], qwq - t[dfn[x]]);
    t[dfn[x]] = qwq;
    for (int i = head[x]; i; i = edge[i].nxt) {
      int v = edge[i].to;
      if (v == fa[x]) continue;
      qwq = calc(a[x], a[v]);
      add(dfn[v], qwq - t[dfn[v]]);
      t[dfn[v]] = qwq;
    }
  }
  long long Query(int x, int y) {
    long long ans = 0;
    while (top[x] != top[y]) {
      if (dep[top[x]] < dep[top[y]]) {
        swap(x, y);
      }
      if (x != top[x]) ans += query(dfn[x]) - query(dfn[top[x]]);
      t[dfn[top[x]]] = calc(a[top[x]], a[fa[top[x]]]);
      ans += t[dfn[top[x]]];
      x = fa[top[x]];
    }
    if (dep[x] > dep[y]) {
      swap(x, y);
    }
    if (x != y) ans += query(dfn[y]) - query(dfn[x]);
    return ans;
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      cin >> u >> v;
      add_edge(u, v);
      add_edge(v, u);
    }
    dfs1(1, 0);
    dfs2(1, 0);
    for (int i = 1; i <= n; i++) add(i, t[i]);
    for (int i = 1; i <= m; i++) {
      int opt, x, y;
      cin >> opt >> x >> y;
      if (opt == 1)
        modify(x, y);
      else
        cout << Query(x, y) << endl;
    }
    return 0;
  }
  