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 MOD = (int)1e9 + 7;
  const int FFTMOD = 119 << 23 | 1;
  const int INF = (int)1e9 + 23111992;
  const long long LINF = (long long)1e18 + 23111992;
  const long double PI = acos((long double)-1);
  const long double EPS = 1e-9;
  inline long long gcd(long long a, long long b) {
    long long r;
    while (b) {
      r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  inline long long fpow(long long n, long long k, int p = MOD) {
    long long r = 1;
    for (; k; k >>= 1) {
      if (k & 1) r = r * n % p;
      n = n * n % p;
    }
    return r;
  }
  template <class T>
  inline int chkmin(T& a, const T& val) {
    return val < a ? a = val, 1 : 0;
  }
  template <class T>
  inline int chkmax(T& a, const T& val) {
    return a < val ? a = val, 1 : 0;
  }
  inline unsigned long long isqrt(unsigned long long k) {
    unsigned long long r = sqrt(k) + 1;
    while (r * r > k) r--;
    return r;
  }
  inline long long icbrt(long long k) {
    long long r = cbrt(k) + 1;
    while (r * r * r > k) r--;
    return r;
  }
  inline void addmod(int& a, int val, int p = MOD) {
    if ((a = (a + val)) >= p) a -= p;
  }
  inline void submod(int& a, int val, int p = MOD) {
    if ((a = (a - val)) < 0) a += p;
  }
  inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
  inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
  inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
  inline int sign(long double x, long double y) { return sign(x - y); }
  mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
  inline int mrand() { return abs((int)mt()); }
  inline int mrand(int k) { return abs((int)mt()) % k; }
  namespace HLD {
  int n;
  vector<vector<int>> adj;
  vector<int> size;
  vector<int> lev;
  vector<int> p;
  vector<int> heavy;
  vector<int> num;
  vector<int> head;
  int cnt;
  void firstdfs(int u, int dad) {
    size[u] = 1, heavy[u] = -1;
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i];
      if (v != dad) {
        p[v] = u, lev[v] = lev[u] + 1;
        firstdfs(v, u);
        size[u] += size[v];
        if (heavy[u] == -1 || size[heavy[u]] < size[v]) {
          heavy[u] = v;
        }
      }
    }
  }
  void dfs(int u, int h, int p) {
    num[u] = cnt++, head[u] = h;
    if (~heavy[u]) {
      dfs(heavy[u], h, u);
    }
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i];
      if (v != p && v != heavy[u]) {
        dfs(v, v, u);
      }
    }
  }
  void build(const vector<vector<int>>& _adj) {
    n = (int)_adj.size();
    adj = _adj;
    size.resize(n);
    lev.resize(n);
    p.resize(n);
    heavy.resize(n);
    num.resize(n);
    head.resize(n);
    cnt = 0;
    firstdfs(0, -1), dfs(0, 0, -1);
  }
  vector<pair<int, int>> getsegs(int u, int v) {
    vector<pair<int, int>> res;
    int hu = head[u], hv = head[v];
    while (hu != hv) {
      if (lev[hu] < lev[hv]) swap(u, v), swap(hu, hv);
      res.push_back(make_pair(num[hu], num[u]));
      u = p[hu], hu = head[u];
    }
    if (lev[u] > lev[v]) swap(u, v);
    res.push_back(make_pair(num[u], num[v]));
    return res;
  }
  }  // namespace HLD
  void chemthan() {
    int n, q;
    cin >> n >> q;
    vector<int> a(n);
    for (int i = (0); i < (n); ++i) cin >> a[i], a[i] = abs(a[i]);
    vector<vector<int>> adj(n);
    map<pair<int, int>, int> hs;
    for (int i = (0); i < (n - 1); ++i) {
      int u, v;
      cin >> u >> v;
      u--, v--;
      adj[u].push_back(v), adj[v].push_back(u);
      hs[make_pair(u, v)] = hs[make_pair(v, u)] = 1;
    }
    HLD::build(adj);
    vector<long long> fen(n + 1);
    auto upd = [&](int p, int v) {
      p++;
      for (; p < int((fen).size()); p += p & -p) {
        fen[p] += v;
      }
    };
    auto query = [&](int p) {
      p++;
      long long res = 0;
      for (; 0 < p; p -= p & -p) {
        res += fen[p];
      }
      return res;
    };
    for (int i = (0); i < (n); ++i) {
      upd(HLD::num[i], a[i]);
    }
    while (q--) {
      int op, u, v;
      cin >> op >> u >> v;
      if (op == 1) {
        u--, v = abs(v);
        upd(HLD::num[u], -a[u]);
        a[u] = v;
        upd(HLD::num[u], a[u]);
      } else {
        u--, v--;
        if (u == v) {
          cout << 0 << "\n";
          continue;
        }
        long long res = 0;
        for (auto [l, r] : HLD::getsegs(u, v)) {
          res += query(r);
          res -= query(l - 1);
        }
        if (!hs[make_pair(u, v)]) {
          res *= 2;
          res -= a[u];
          res -= a[v];
        }
        cout << res << "\n";
      }
    }
  }
  int32_t main(int32_t argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
      assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
      assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
  }
  