#include <bits/stdc++.h>
using namespace std;
const int Nmax = 1e5 + 10;
int n, m;
vector<int> e[Nmax];
int par[Nmax][20];
int in[Nmax], out[Nmax], ti;
void dfs(int u, int p) {
  in[u] = ++ti;
  par[u][0] = p;
  for (int i = 1; i < 20; ++i) par[u][i] = par[par[u][i - 1]][i - 1];
  for (auto v : e[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
  out[u] = ti;
}
long long BIT[Nmax];
void add(int id, int delta) {
  for (; id <= n; id += id & (-id)) BIT[id] += delta;
}
long long get(int id) {
  long long ans = 0;
  for (; id; id -= id & (-id)) ans += BIT[id];
  return ans;
}
int check(int u, int v) { return in[u] <= in[v] && out[v] <= out[u]; }
int LCA(int u, int v) {
  if (check(u, v)) return u;
  for (int i = 19; i >= 0; --i)
    if (par[u][i] && !check(par[u][i], v)) u = par[u][i];
  return par[u][0];
}
int a[Nmax];
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];
    a[i] = abs(a[i]);
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) add(in[i], a[i]), add(out[i] + 1, -a[i]);
  while (m--) {
    int type;
    cin >> type;
    if (type == 1) {
      int u, c;
      cin >> u >> c;
      c = abs(c);
      add(in[u], c - a[u]);
      add(out[u] + 1, a[u] - c);
      a[u] = c;
    } else {
      int u, v;
      cin >> u >> v;
      int w = LCA(u, v);
      cout << (get(in[u]) + get(in[v]) - 2 * get(in[w]) + a[w]) * 2 - a[u] -
                  a[v]
           << '\n';
    }
  }
}

### Prompt

In Cpp, your task is to solve 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;
const int Nmax = 1e5 + 10;
int n, m;
vector<int> e[Nmax];
int par[Nmax][20];
int in[Nmax], out[Nmax], ti;
void dfs(int u, int p) {
  in[u] = ++ti;
  par[u][0] = p;
  for (int i = 1; i < 20; ++i) par[u][i] = par[par[u][i - 1]][i - 1];
  for (auto v : e[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
  out[u] = ti;
}
long long BIT[Nmax];
void add(int id, int delta) {
  for (; id <= n; id += id & (-id)) BIT[id] += delta;
}
long long get(int id) {
  long long ans = 0;
  for (; id; id -= id & (-id)) ans += BIT[id];
  return ans;
}
int check(int u, int v) { return in[u] <= in[v] && out[v] <= out[u]; }
int LCA(int u, int v) {
  if (check(u, v)) return u;
  for (int i = 19; i >= 0; --i)
    if (par[u][i] && !check(par[u][i], v)) u = par[u][i];
  return par[u][0];
}
int a[Nmax];
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];
    a[i] = abs(a[i]);
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) add(in[i], a[i]), add(out[i] + 1, -a[i]);
  while (m--) {
    int type;
    cin >> type;
    if (type == 1) {
      int u, c;
      cin >> u >> c;
      c = abs(c);
      add(in[u], c - a[u]);
      add(out[u] + 1, a[u] - c);
      a[u] = c;
    } else {
      int u, v;
      cin >> u >> v;
      int w = LCA(u, v);
      cout << (get(in[u]) + get(in[v]) - 2 * get(in[w]) + a[w]) * 2 - a[u] -
                  a[v]
           << '\n';
    }
  }
}

```