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;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    T w = 1, ch = getchar();
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ '0'), ch = getchar();
    x = x * w;
  }
  const long long maxn = 200000 + 10;
  long long n, m, r, mod;
  long long e, beg[maxn], nex[maxn], to[maxn], w[maxn], wt[maxn];
  long long a[maxn << 2], laz[maxn << 2];
  long long son[maxn], id[maxn], fa[maxn], cnt, dep[maxn], siz[maxn], top[maxn];
  long long res = 0;
  inline void add(long long x, long long y) {
    to[++e] = y;
    nex[e] = beg[x];
    beg[x] = e;
  }
  inline void pushdown(long long rt, long long lenn) {
    laz[rt << 1] += laz[rt];
    laz[rt << 1 | 1] += laz[rt];
    a[rt << 1] += laz[rt] * (lenn - (lenn >> 1));
    a[rt << 1 | 1] += laz[rt] * (lenn >> 1);
    a[rt << 1] %= mod;
    a[rt << 1 | 1] %= mod;
    laz[rt] = 0;
  }
  inline void build(long long rt, long long l, long long r) {
    if (l == r) {
      a[rt] = wt[l];
      if (a[rt] > mod) a[rt] %= mod;
      return;
    }
    build(rt << 1, l, ((l + r) >> 1));
    build(rt << 1 | 1, ((l + r) >> 1) + 1, r);
    a[rt] = (a[rt << 1] + a[rt << 1 | 1]) % mod;
  }
  inline void query(long long rt, long long l, long long r, long long L,
                    long long R) {
    if (L <= l && r <= R) {
      res += a[rt];
      res %= mod;
      return;
    } else {
      if (laz[rt]) pushdown(rt, (r - l + 1));
      if (L <= ((l + r) >> 1)) query(rt << 1, l, ((l + r) >> 1), L, R);
      if (R > ((l + r) >> 1)) query(rt << 1 | 1, ((l + r) >> 1) + 1, r, L, R);
    }
  }
  inline void update(long long rt, long long l, long long r, long long L,
                     long long R, long long k) {
    if (L <= l && r <= R) {
      laz[rt] += k;
      a[rt] += k * (r - l + 1);
    } else {
      if (laz[rt]) pushdown(rt, (r - l + 1));
      if (L <= ((l + r) >> 1)) update(rt << 1, l, ((l + r) >> 1), L, R, k);
      if (R > ((l + r) >> 1)) update(rt << 1 | 1, ((l + r) >> 1) + 1, r, L, R, k);
      a[rt] = (a[rt << 1] + a[rt << 1 | 1]) % mod;
    }
  }
  inline long long qRange(long long x, long long y) {
    long long ans = 0;
    while (top[x] != top[y]) {
      if (dep[top[x]] < dep[top[y]]) swap(x, y);
      res = 0;
      query(1, 1, n, id[top[x]], id[x]);
      ans += res;
      ans %= mod;
      x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    res = 0;
    query(1, 1, n, id[x], id[y]);
    ans += res;
    return ans % mod;
  }
  inline void updRange(long long x, long long y, long long k) {
    k %= mod;
    while (top[x] != top[y]) {
      if (dep[top[x]] < dep[top[y]]) swap(x, y);
      update(1, 1, n, id[top[x]], id[x], k);
      x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    update(1, 1, n, id[x], id[y], k);
  }
  inline long long qSon(long long x) {
    res = 0;
    query(1, 1, n, id[x], id[x] + siz[x] - 1);
    return res;
  }
  inline void updSon(long long x, long long k) {
    update(1, 1, n, id[x], id[x] + siz[x] - 1, k);
  }
  inline void dfs1(long long x, long long f, long long deep) {
    dep[x] = deep;
    fa[x] = f;
    siz[x] = 1;
    long long maxson = -1;
    for (long long i = beg[x]; i; i = nex[i]) {
      long long y = to[i];
      if (y == f) continue;
      dfs1(y, x, deep + 1);
      siz[x] += siz[y];
      if (siz[y] > maxson) son[x] = y, maxson = siz[y];
    }
  }
  inline void dfs2(long long x, long long topf) {
    id[x] = ++cnt;
    wt[cnt] = w[x];
    top[x] = topf;
    if (!son[x]) return;
    dfs2(son[x], topf);
    for (long long i = beg[x]; i; i = nex[i]) {
      long long y = to[i];
      if (y == fa[x] || y == son[x]) continue;
      dfs2(y, y);
    }
  }
  signed main() {
    read(n);
    read(m);
    r = 1, mod = 1e18;
    for (long long i = 1; i <= n; i++) {
      read(w[i]);
      if (w[i] < 0) w[i] = -w[i];
    }
    for (long long i = 1; i < n; i++) {
      long long a, b;
      read(a);
      read(b);
      add(a, b);
      add(b, a);
    }
    dfs1(r, 0, 1);
    dfs2(r, r);
    build(1, 1, n);
    while (m--) {
      long long k, x, y, z;
      read(k);
      if (k == 1) {
        read(x);
        read(y);
        if (y < 0) y = -y;
        updRange(x, x, y - w[x]);
        w[x] = y;
      } else if (k == 2) {
        read(x);
        read(y);
        printf("%lld\n", qRange(x, y) * 2 - w[x] - w[y]);
      }
    }
  }
  