#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 50;
int n, q, dfn[N], rr[N], cnt, f[N][20], d[N], a[N];
long long bit[N];
vector<int> v[N];
void dfs(int x, int ff) {
  dfn[x] = ++cnt;
  f[x][0] = ff;
  d[x] = d[ff] + 1;
  for (int i = 0; i < 18; i++) f[x][i + 1] = f[f[x][i]][i];
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (y == ff) continue;
    dfs(y, x);
  }
  rr[x] = cnt;
}
int lca(int x, int y) {
  if (d[x] < d[y]) swap(x, y);
  for (int i = 18; ~i; i--)
    if (d[f[x][i]] >= d[y]) x = f[x][i];
  if (x == y) return x;
  for (int i = 18; ~i; i--)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
void add(int x, int y) {
  for (; x <= n; x += x & -x) bit[x] += y;
}
long long ask(int x) {
  long long z = 0;
  for (; x; x -= x & -x) z += bit[x];
  return z;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] = abs(a[i]);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) add(dfn[i], a[i]), add(rr[i] + 1, -a[i]);
  for (int i = 1, op, x, y; i <= q; i++) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      add(dfn[x], -a[x]);
      add(rr[x] + 1, a[x]);
      add(dfn[x], a[x] = abs(y));
      add(rr[x] + 1, -a[x]);
    } else {
      int z = lca(x, y), zz = f[z][0];
      printf("%lld\n",
             2 * (ask(dfn[x]) + ask(dfn[y]) - ask(dfn[z]) - ask(dfn[zz])) -
                 a[x] - a[y]);
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 1e5 + 50;
int n, q, dfn[N], rr[N], cnt, f[N][20], d[N], a[N];
long long bit[N];
vector<int> v[N];
void dfs(int x, int ff) {
  dfn[x] = ++cnt;
  f[x][0] = ff;
  d[x] = d[ff] + 1;
  for (int i = 0; i < 18; i++) f[x][i + 1] = f[f[x][i]][i];
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (y == ff) continue;
    dfs(y, x);
  }
  rr[x] = cnt;
}
int lca(int x, int y) {
  if (d[x] < d[y]) swap(x, y);
  for (int i = 18; ~i; i--)
    if (d[f[x][i]] >= d[y]) x = f[x][i];
  if (x == y) return x;
  for (int i = 18; ~i; i--)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
void add(int x, int y) {
  for (; x <= n; x += x & -x) bit[x] += y;
}
long long ask(int x) {
  long long z = 0;
  for (; x; x -= x & -x) z += bit[x];
  return z;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] = abs(a[i]);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) add(dfn[i], a[i]), add(rr[i] + 1, -a[i]);
  for (int i = 1, op, x, y; i <= q; i++) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      add(dfn[x], -a[x]);
      add(rr[x] + 1, a[x]);
      add(dfn[x], a[x] = abs(y));
      add(rr[x] + 1, -a[x]);
    } else {
      int z = lca(x, y), zz = f[z][0];
      printf("%lld\n",
             2 * (ask(dfn[x]) + ask(dfn[y]) - ask(dfn[z]) - ask(dfn[zz])) -
                 a[x] - a[y]);
    }
  }
  return 0;
}

```