#include <bits/stdc++.h>
using namespace std;
vector<int> v[100005];
int n, m, a[100005], dfn[100005], f[18][100005], T, dfr[100005], d[100005];
long long t[100005];
void dfs(int p, int fa) {
  f[0][p] = fa;
  dfn[p] = ++T;
  d[p] = d[fa] + 1;
  for (int i = 1; i < 18; ++i) f[i][p] = f[i - 1][f[i - 1][p]];
  for (int x : v[p])
    if (x != fa) dfs(x, p);
  dfr[p] = T + 1;
}
void add(int p, int v) {
  for (; p < 100005; p += p & -p) t[p] += v;
}
long long ask(int p, long long a = 0) {
  for (; p; p -= p & -p) a += t[p];
  return a;
}
int lca(int x, int y) {
  if (d[x] < d[y]) swap(x, y);
  for (int i = 17; ~i; --i)
    if (d[x] - d[y] >> i & 1) x = f[i][x];
  if (x == y) return x;
  for (int i = 17; ~i; --i)
    if (f[i][x] != f[i][y]) x = f[i][x], y = f[i][y];
  return f[0][x];
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), a[i] = abs(a[i]);
  for (int i = 1, a, b; i < n; ++i)
    scanf("%d%d", &a, &b), v[a].push_back(b), v[b].push_back(a);
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) add(dfn[i], a[i]), add(dfr[i], -a[i]);
  for (int i = 1; i <= m; ++i) {
    int o, x, y;
    scanf("%d%d%d", &o, &x, &y);
    y = abs(y);
    if (o == 1)
      add(dfn[x], y - a[x]), add(dfr[x], a[x] - y), a[x] = y;
    else
      printf("%lld\n", (ask(dfn[x]) + ask(dfn[y]) - ask(dfn[lca(x, y)]) -
                        ask(dfn[f[0][lca(x, y)]])) *
                               2 -
                           a[x] - a[y]);
  }
}

### Prompt

Create a solution in CPP for 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;
vector<int> v[100005];
int n, m, a[100005], dfn[100005], f[18][100005], T, dfr[100005], d[100005];
long long t[100005];
void dfs(int p, int fa) {
  f[0][p] = fa;
  dfn[p] = ++T;
  d[p] = d[fa] + 1;
  for (int i = 1; i < 18; ++i) f[i][p] = f[i - 1][f[i - 1][p]];
  for (int x : v[p])
    if (x != fa) dfs(x, p);
  dfr[p] = T + 1;
}
void add(int p, int v) {
  for (; p < 100005; p += p & -p) t[p] += v;
}
long long ask(int p, long long a = 0) {
  for (; p; p -= p & -p) a += t[p];
  return a;
}
int lca(int x, int y) {
  if (d[x] < d[y]) swap(x, y);
  for (int i = 17; ~i; --i)
    if (d[x] - d[y] >> i & 1) x = f[i][x];
  if (x == y) return x;
  for (int i = 17; ~i; --i)
    if (f[i][x] != f[i][y]) x = f[i][x], y = f[i][y];
  return f[0][x];
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), a[i] = abs(a[i]);
  for (int i = 1, a, b; i < n; ++i)
    scanf("%d%d", &a, &b), v[a].push_back(b), v[b].push_back(a);
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) add(dfn[i], a[i]), add(dfr[i], -a[i]);
  for (int i = 1; i <= m; ++i) {
    int o, x, y;
    scanf("%d%d%d", &o, &x, &y);
    y = abs(y);
    if (o == 1)
      add(dfn[x], y - a[x]), add(dfr[x], a[x] - y), a[x] = y;
    else
      printf("%lld\n", (ask(dfn[x]) + ask(dfn[y]) - ask(dfn[lca(x, y)]) -
                        ask(dfn[f[0][lca(x, y)]])) *
                               2 -
                           a[x] - a[y]);
  }
}

```