#include <bits/stdc++.h>
using namespace std;
class Fenwick {
  vector<long long> tree;

 public:
  Fenwick() {}
  Fenwick(int n, long long init) : tree(n + 5, 0) {}
  void add(int idx, long long val) {
    for (int i = idx; i < (int)tree.size(); i += i & -i) {
      tree[i] += val;
    }
  }
  long long get(int idx) {
    long long ans = 0;
    for (int i = idx; i > 0; i -= i & -i) {
      ans += tree[i];
    }
    return ans;
  }
  long long query(int l, int r) {
    long long ans = get(r) - get(l - 1);
    return ans;
  }
  void set(int idx, long long val) { add(idx, val - query(idx, idx)); }
};
template <typename T, class DS>
class HLD {
  int vertexc;
  vector<int> *adj;
  vector<int> subtree_size;
  DS structure;
  DS aux;
  void build_sizes(int vertex, int parent) {
    subtree_size[vertex] = 1;
    for (int child : adj[vertex]) {
      if (child != parent) {
        build_sizes(child, vertex);
        subtree_size[vertex] += subtree_size[child];
      }
    }
  }
  int cur;
  vector<int> ord;
  vector<int> chain_root;
  vector<int> par;
  void build_hld(int vertex, int parent, int chain_source) {
    cur++;
    ord[vertex] = cur;
    chain_root[vertex] = chain_source;
    par[vertex] = parent;
    if (adj[vertex].size() > 1 || (vertex == 1 && adj[vertex].size() == 1)) {
      int big_child, big_size = -1;
      for (int child : adj[vertex]) {
        if ((child != parent) && (subtree_size[child] > big_size)) {
          big_child = child;
          big_size = subtree_size[child];
        }
      }
      build_hld(big_child, vertex, chain_source);
      for (int child : adj[vertex]) {
        if ((child != parent) && (child != big_child))
          build_hld(child, vertex, child);
      }
    }
  }

 public:
  HLD(int _vertexc) {
    vertexc = _vertexc;
    adj = new vector<int>[vertexc + 5];
  }
  void add_edge(int u, int v) {
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  void build(T initial) {
    subtree_size = vector<int>(vertexc + 5);
    ord = vector<int>(vertexc + 5);
    chain_root = vector<int>(vertexc + 5);
    par = vector<int>(vertexc + 5);
    cur = 0;
    build_sizes(1, -1);
    build_hld(1, -1, 1);
    structure = DS(vertexc + 5, initial);
    aux = DS(50, initial);
  }
  void set(int vertex, int value) { structure.set(ord[vertex], value); }
  T query_path(int u, int v) {
    int cur_id = 0;
    while (chain_root[u] != chain_root[v]) {
      if (ord[u] > ord[v]) {
        cur_id++;
        aux.set(cur_id, structure.query(ord[chain_root[u]], ord[u]));
        u = par[chain_root[u]];
      } else {
        cur_id++;
        aux.set(cur_id, structure.query(ord[chain_root[v]], ord[v]));
        v = par[chain_root[v]];
      }
    }
    cur_id++;
    aux.set(cur_id, structure.query(min(ord[u], ord[v]), max(ord[u], ord[v])));
    return aux.query(1, cur_id);
  }
  void print() {
    for (int i = 1; i <= vertexc; i++)
      cout << i << ' ' << ord[i] << ' ' << chain_root[i] << ' ' << par[i]
           << endl;
  }
};
const int MAX_N = 1e5 + 5;
long long arr[MAX_N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, qc;
  cin >> n >> qc;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    arr[i] = abs(arr[i]);
  }
  HLD<long long, Fenwick> hld(n + 5);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    hld.add_edge(u, v);
  }
  hld.build(1);
  for (int i = 1; i <= n; i++) {
    hld.set(i, arr[i]);
  }
  for (int i = 0; i < qc; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int u, c;
      cin >> u >> c;
      arr[u] = abs(c);
      hld.set(u, arr[u]);
    } else {
      int u, v;
      cin >> u >> v;
      long long ans = hld.query_path(u, v);
      ans *= 2;
      ans -= arr[u];
      ans -= arr[v];
      cout << ans << '\n';
    }
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
class Fenwick {
  vector<long long> tree;

 public:
  Fenwick() {}
  Fenwick(int n, long long init) : tree(n + 5, 0) {}
  void add(int idx, long long val) {
    for (int i = idx; i < (int)tree.size(); i += i & -i) {
      tree[i] += val;
    }
  }
  long long get(int idx) {
    long long ans = 0;
    for (int i = idx; i > 0; i -= i & -i) {
      ans += tree[i];
    }
    return ans;
  }
  long long query(int l, int r) {
    long long ans = get(r) - get(l - 1);
    return ans;
  }
  void set(int idx, long long val) { add(idx, val - query(idx, idx)); }
};
template <typename T, class DS>
class HLD {
  int vertexc;
  vector<int> *adj;
  vector<int> subtree_size;
  DS structure;
  DS aux;
  void build_sizes(int vertex, int parent) {
    subtree_size[vertex] = 1;
    for (int child : adj[vertex]) {
      if (child != parent) {
        build_sizes(child, vertex);
        subtree_size[vertex] += subtree_size[child];
      }
    }
  }
  int cur;
  vector<int> ord;
  vector<int> chain_root;
  vector<int> par;
  void build_hld(int vertex, int parent, int chain_source) {
    cur++;
    ord[vertex] = cur;
    chain_root[vertex] = chain_source;
    par[vertex] = parent;
    if (adj[vertex].size() > 1 || (vertex == 1 && adj[vertex].size() == 1)) {
      int big_child, big_size = -1;
      for (int child : adj[vertex]) {
        if ((child != parent) && (subtree_size[child] > big_size)) {
          big_child = child;
          big_size = subtree_size[child];
        }
      }
      build_hld(big_child, vertex, chain_source);
      for (int child : adj[vertex]) {
        if ((child != parent) && (child != big_child))
          build_hld(child, vertex, child);
      }
    }
  }

 public:
  HLD(int _vertexc) {
    vertexc = _vertexc;
    adj = new vector<int>[vertexc + 5];
  }
  void add_edge(int u, int v) {
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  void build(T initial) {
    subtree_size = vector<int>(vertexc + 5);
    ord = vector<int>(vertexc + 5);
    chain_root = vector<int>(vertexc + 5);
    par = vector<int>(vertexc + 5);
    cur = 0;
    build_sizes(1, -1);
    build_hld(1, -1, 1);
    structure = DS(vertexc + 5, initial);
    aux = DS(50, initial);
  }
  void set(int vertex, int value) { structure.set(ord[vertex], value); }
  T query_path(int u, int v) {
    int cur_id = 0;
    while (chain_root[u] != chain_root[v]) {
      if (ord[u] > ord[v]) {
        cur_id++;
        aux.set(cur_id, structure.query(ord[chain_root[u]], ord[u]));
        u = par[chain_root[u]];
      } else {
        cur_id++;
        aux.set(cur_id, structure.query(ord[chain_root[v]], ord[v]));
        v = par[chain_root[v]];
      }
    }
    cur_id++;
    aux.set(cur_id, structure.query(min(ord[u], ord[v]), max(ord[u], ord[v])));
    return aux.query(1, cur_id);
  }
  void print() {
    for (int i = 1; i <= vertexc; i++)
      cout << i << ' ' << ord[i] << ' ' << chain_root[i] << ' ' << par[i]
           << endl;
  }
};
const int MAX_N = 1e5 + 5;
long long arr[MAX_N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, qc;
  cin >> n >> qc;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    arr[i] = abs(arr[i]);
  }
  HLD<long long, Fenwick> hld(n + 5);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    hld.add_edge(u, v);
  }
  hld.build(1);
  for (int i = 1; i <= n; i++) {
    hld.set(i, arr[i]);
  }
  for (int i = 0; i < qc; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int u, c;
      cin >> u >> c;
      arr[u] = abs(c);
      hld.set(u, arr[u]);
    } else {
      int u, v;
      cin >> u >> v;
      long long ans = hld.query_path(u, v);
      ans *= 2;
      ans -= arr[u];
      ans -= arr[v];
      cout << ans << '\n';
    }
  }
}

```