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;
  bool debug;
  template <typename T>
  struct TreePathQuery {
    vector<int> tin, tout;
    int n, root;
    vector<T> nodeVal;
    vector<int> depth;
    vector<vector<int>> adj;
    TreePathQuery(int n, int root, T (*op)(const T &a, const T &b), T sentinel)
        : n(n), root(root), adj(n), SENTINEL(sentinel), depth(n) {
      this->op = op;
    }
    void addEdge(int u, int v) {
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    T queryEx(int a, int b) {
      assert(isBuilt);
      if (a == b) return SENTINEL;
      if (isAncestor(a, b)) return dist(b, a);
      if (isAncestor(b, a)) return dist(a, b);
      int ca = lca(a, b);
      return op(dist(a, ca), dist(b, ca));
    }
    T queryIn(int a, int b) {
      assert(isBuilt);
      if (a == b) return nodeVal[a];
      if (isAncestor(a, b)) return op(nodeVal[a], dist(b, a));
      if (isAncestor(b, a)) return op(nodeVal[b], dist(a, b));
      int ca = lca(a, b);
      return op(nodeVal[ca], op(dist(a, ca), dist(b, ca)));
    }
    int nthAncestor(int u, int nth) {
      assert(isBuilt);
      for (int i = 0; nth; ++i) {
        if (nth & 1) {
          u = anc[u][i];
        }
        nth >>= 1;
      }
      return u;
    }
    int lca(int a, int b) {
      assert(isBuilt);
      if (isAncestor(a, b)) return a;
      if (isAncestor(b, a)) return b;
      for (int i = l; i >= 0; i--)
        if (!isAncestor(anc[a][i], b)) a = anc[a][i];
      return anc[a][0];
    }
    void build(vector<T> &nodeVal) {
      this->nodeVal = nodeVal;
      assert(nodeVal.size() == n);
      isBuilt = true;
      tin.resize(n);
      tout.resize(n);
      timer = 0;
      l = ceil(log2(n));
      anc.assign(n, vector<int>(l + 1));
      dp.assign(n, vector<T>(l + 1));
      dfs(root, root, 0);
    }
    bool isAncestor(int a, int b) {
      return assert(isBuilt), tin[a] <= tin[b] && tout[a] >= tout[b];
    }
  
   private:
    int l, timer;
    bool isBuilt = false;
    T SENTINEL = 0;
    vector<vector<int>> anc;
    vector<vector<T>> dp;
    T(*op)
    (const T &a, const T &b);
    void dfs(int u, int p, int dep) {
      depth[u] = dep;
      tin[u] = ++timer;
      anc[u][0] = p;
      dp[u][0] = nodeVal[u];
      for (int i = 1; i <= l; ++i) {
        anc[u][i] = anc[anc[u][i - 1]][i - 1];
        dp[u][i] = op(dp[anc[u][i - 1]][i - 1], dp[u][i - 1]);
      }
      for (int v : adj[u])
        if (v != p) dfs(v, u, dep + 1);
      tout[u] = ++timer;
    }
    T dist(int u, int ancestor) {
      if (u == ancestor) return nodeVal[u];
      T sum = SENTINEL;
      for (int i = l; i >= 0; i--)
        if (!isAncestor(anc[u][i], ancestor)) {
          sum = op(sum, dp[u][i]);
          u = anc[u][i];
        }
      sum = op(sum, dp[u][0]);
      return sum;
    }
  };
  template <typename T>
  T opSum(const T &a, const T &b) {
    return a + b;
  }
  template <typename T>
  T opMin(const T &a, const T &b) {
    return min(a, b);
  }
  template <typename T>
  T opMax(const T &a, const T &b) {
    return max(a, b);
  }
  template <typename T>
  T opGcd(const T &a, const T &b) {
    return __gcd(a, b);
  }
  template <typename T>
  struct FenwickTreeSum {
   public:
    vector<T> ft;
    FenwickTreeSum(int n) { ft.assign(n + 1, 0); }
    FenwickTreeSum(vector<T> arr) {
      int n = arr.size();
      ft.assign(n + 1, 0);
      for (int i = 0; i < n; i++) update(i, arr[i]);
    }
    T query(int b) {
      b++;
      T sum = 0;
      for (; b; b -= LSOne(b)) sum = op(sum, ft[b]);
      return sum;
    }
    T rangeQuery(int a, int b) {
      return invOp(query(b), (a == 0 ? 0 : query(a - 1)));
    }
    void update(int k, T v) {
      k++;
      for (; k < ft.size(); k += LSOne(k)) ft[k] = op(ft[k], v);
    }
  
   private:
    int LSOne(int x) { return x & (-x); }
    T op(const T &a, const T &b) { return a + b; }
    T invOp(const T &a, const T &b) { return a - b; }
  };
  template <typename T>
  struct SqrtLCASum {
   public:
    SqrtLCASum(const vector<T> &v)
        : n((int)v.size()), lg(log2Up(n)), v(v), clz(1 << lg), onLayer(lg + 1) {
      clz[0] = 0;
      for (int i = 1; i < (int)clz.size(); i++) clz[i] = clz[i >> 1] + 1;
      int tlg = lg;
      while (tlg > 1) {
        onLayer[tlg] = (int)layers.size();
        layers.push_back(tlg);
        tlg = (tlg + 1) >> 1;
      }
      for (int i = lg - 1; i >= 0; i--)
        onLayer[i] = max(onLayer[i], onLayer[i + 1]);
      pref.assign(layers.size(), vector<T>(n));
      suf.assign(layers.size(), vector<T>(n));
      between.assign(layers.size(), vector<T>(1 << lg));
      build(0, 0, n);
    }
    inline T query(int l, int r) {
      if (l == r) return v[l];
      if (l + 1 == r) return op(v[l], v[r]);
      int layer = onLayer[clz[l ^ r]];
      int bSzLog = (layers[layer] + 1) >> 1;
      int bCntLog = layers[layer] >> 1;
      int lBound = (l >> layers[layer]) << layers[layer];
      int lBlock = ((l - lBound) >> bSzLog) + 1;
      int rBlock = ((r - lBound) >> bSzLog) - 1;
      T ans = suf[layer][l];
      if (lBlock <= rBlock) {
        ans = op(ans, between[layer][lBound + (lBlock << bCntLog) + rBlock]);
      }
      ans = op(ans, pref[layer][r]);
      return ans;
    }
  
   private:
    int n, lg;
    vector<T> v, clz, layers, onLayer;
    vector<vector<T>> pref, suf, between;
    void build(int layer, int lBound, int rBound) {
      if (layer >= (int)layers.size()) return;
      int bSzLog = (layers[layer] + 1) >> 1;
      int bCntLog = layers[layer] >> 1;
      int bSz = 1 << bSzLog;
      int bCnt = 0;
      for (int l = lBound; l < rBound; l += bSz) {
        bCnt++;
        int r = min(l + bSz, rBound);
        pref[layer][l] = v[l];
        for (int i = l + 1; i < r; i++)
          pref[layer][i] = op(pref[layer][i - 1], v[i]);
        suf[layer][r - 1] = v[r - 1];
        for (int i = r - 2; i >= l; i--)
          suf[layer][i] = op(v[i], suf[layer][i + 1]);
        build(layer + 1, l, r);
      }
      for (int i = 0; i < bCnt; i++) {
        T ans;
        for (int j = i; j < bCnt; j++) {
          T add = suf[layer][lBound + (j << bSzLog)];
          ans = (i == j) ? add : op(ans, add);
          between[layer][lBound + (i << bCntLog) + j] = ans;
        }
      }
    }
    inline int log2Up(int n) {
      int res = 0;
      while ((1 << res) < n) res++;
      return res;
    }
    T op(const T &a, const T &b) { return min(a, b); }
  };
  template <typename T, typename RSQ>
  struct TreePathQuerySum {
   public:
    int n, root;
    vector<T> nodeVal;
    TreePathQuerySum(vector<T> nodeVal, int root)
        : n(nodeVal.size()),
          root(root),
          adj(n),
          first(n),
          inOrder(n),
          last(n),
          revOrder(n),
          isBuilt(false) {
      this->nodeVal = nodeVal;
      euler.reserve(2 * n);
    }
    void addEdge(int u, int v) {
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    int getLCA(int a, int b) {
      if (!isBuilt) {
        build();
      }
      int l = first[a], r = first[b];
      if (l > r) swap(l, r);
      return revOrder[sqt->query(l, r)];
    }
    T queryIn(int a, int b) {
      int lca = getLCA(a, b);
      T lKeA = rsq->rangeQuery(first[lca], first[a]);
      T lKeB = rsq->rangeQuery(first[lca], first[b]);
      return lKeA - nodeVal[lca] + lKeB;
    }
    T queryEx(int a, int b) {
      int lca = getLCA(a, b);
      T lKeA = rsq->rangeQuery(first[lca], first[a]);
      T lKeB = rsq->rangeQuery(first[lca], first[b]);
      return lKeA - 2 * nodeVal[lca] + lKeB;
    }
    void update(int u, T val) {
      nodeVal[u] += val;
      rsq->update(first[u], val);
      rsq->update(last[u] + 1, -val);
    }
    void set(int u, T val) {
      T diff = val - nodeVal[u];
      nodeVal[u] = val;
      rsq->update(first[u], diff);
      rsq->update(last[u] + 1, -diff);
    }
    void build() {
      isBuilt = true;
      visited.assign(n, false);
      wkt = 0;
      rsq = new RSQ(vector<T>(2 * n, 0));
      dfs(root);
      sqt = new SqrtLCASum<int>(euler);
    }
  
   private:
    SqrtLCASum<int> *sqt;
    RSQ *rsq;
    bool isBuilt;
    vector<int> euler, first, inOrder, revOrder, last;
    vector<vector<int>> adj;
    int wkt;
    vector<bool> visited;
    void dfs(int u) {
      visited[u] = true;
      inOrder[u] = wkt++;
      first[u] = euler.size();
      rsq->update(first[u], nodeVal[u]);
      last[u] = first[u];
      revOrder[inOrder[u]] = u;
      euler.push_back(inOrder[u]);
      for (int v : adj[u])
        if (!visited[v]) {
          dfs(v);
          last[u] = euler.size();
          euler.push_back(inOrder[u]);
        }
      rsq->update(last[u] + 1, -nodeVal[u]);
    }
  };
  int main(int argc, char *argv[]) {
    debug = argc > 1;
    int n, q;
    scanf("%d%d", &n, &q);
    TreePathQuery<long long> graph(n, 0, opMax<long long>, 0);
    vector<long long> nodeVal(n);
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &x);
      nodeVal[i] = abs(x);
    }
    vector<pair<int, int>> edges;
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      u--, v--;
      graph.addEdge(u, v);
      edges.emplace_back(u, v);
    }
    graph.build(nodeVal);
    TreePathQuerySum<long long, FenwickTreeSum<long long>> tpq(nodeVal, 0);
    for (auto [a, b] : edges) {
      tpq.addEdge(a, b);
    }
    tpq.build();
    if (debug) {
      cout << "graph.depth"
           << ": ";
      for (auto i : graph.depth) {
        cout << i << ' ';
      }
      cout << endl;
    }
    int c, u, v;
    while (q--) {
      scanf("%d%d%d", &c, &u, &v);
      if (c == 1) {
        tpq.set(u - 1, abs(v));
      } else {
        u--, v--;
        if (debug) {
          cout << "u"
               << ": " << u << endl;
        }
        if (debug) {
          cout << "v"
               << ": " << v << endl;
        }
        int lca = tpq.getLCA(u, v);
        if (debug) {
          cout << "lca"
               << ": " << lca << endl;
        }
        long long ans = -1;
        if (lca == v) {
          swap(u, v);
        }
        if (u == v) {
          ans = 0;
        } else if (lca == u) {
          if (u == graph.nthAncestor(v, 1)) {
            ans = tpq.queryIn(u, v);
            if (debug) {
              cout << "ans"
                   << ": " << ans << endl;
            }
          } else {
            int step = graph.depth[v] - graph.depth[u];
            int parentV = graph.nthAncestor(v, 1);
            int childU = graph.nthAncestor(v, step - 1);
            ans = tpq.queryIn(u, v) + tpq.queryIn(childU, parentV);
          }
        } else {
          ans = tpq.queryIn(u, v) +
                tpq.queryIn(graph.nthAncestor(u, 1), graph.nthAncestor(v, 1));
        }
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  