#include <bits/stdc++.h>
using namespace std;
long long in_time[200005];
long long a[200005];
long long out_time[200005];
long long depth[200005];
long long tim = 1, n;
long long bit[2000005];
vector<long long> adj[200005];
long long par[200005][25];
void upd(long long ind, long long v, long long n) {
  while (ind <= n) {
    bit[ind] += v;
    ind += (ind & (-ind));
  }
}
long long query(long long ind) {
  long long s = 0;
  while (ind > 0) {
    s += bit[ind];
    ind -= (ind & (-ind));
  }
  return s;
}
void dfs(long long v, long long p) {
  depth[v] = depth[p] + 1;
  par[v][0] = p;
  in_time[v] = tim++;
  upd(in_time[v], a[v], 2 * n);
  for (long long i = 1; i <= 20; i++) {
    par[v][i] = par[par[v][i - 1]][i - 1];
  }
  for (auto to : adj[v]) {
    if (to == p) continue;
    dfs(to, v);
  }
  out_time[v] = tim++;
  upd(out_time[v], -a[v], 2 * n);
}
long long lca(long long u, long long v) {
  if (depth[u] > depth[v]) {
    swap(u, v);
  }
  long long diff = abs(depth[u] - depth[v]);
  for (long long i = 0; i <= 20; i++) {
    if (diff & (1ll << i)) {
      v = par[v][i];
    }
  }
  if (u == v) {
    return v;
  }
  for (long long i = 20; i >= 0; i--) {
    if (par[u][i] == par[v][i]) {
      continue;
    }
    u = par[u][i];
    v = par[v][i];
  }
  return par[u][0];
}
void solve(long long tc) {
  long long q, x, y;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  for (long long i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  long long ty;
  while (q--) {
    cin >> ty >> x >> y;
    if (ty == 1) {
      upd(in_time[x], -a[x], 2 * n);
      upd(out_time[x], a[x], 2 * n);
      a[x] = abs(y);
      upd(in_time[x], a[x], 2 * n);
      upd(out_time[x], -a[x], 2 * n);
    } else {
      long long lc = lca(x, y);
      long long d = 2 * (query(in_time[x]) + query(in_time[y]) -
                         2 * query(in_time[lc]) + a[lc]);
      cout << d - a[x] - a[y] << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  long long t = 1;
  long long tc = 1;
  while (t--) {
    solve(tc);
    tc++;
  }
  return 0;
}

### 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;
long long in_time[200005];
long long a[200005];
long long out_time[200005];
long long depth[200005];
long long tim = 1, n;
long long bit[2000005];
vector<long long> adj[200005];
long long par[200005][25];
void upd(long long ind, long long v, long long n) {
  while (ind <= n) {
    bit[ind] += v;
    ind += (ind & (-ind));
  }
}
long long query(long long ind) {
  long long s = 0;
  while (ind > 0) {
    s += bit[ind];
    ind -= (ind & (-ind));
  }
  return s;
}
void dfs(long long v, long long p) {
  depth[v] = depth[p] + 1;
  par[v][0] = p;
  in_time[v] = tim++;
  upd(in_time[v], a[v], 2 * n);
  for (long long i = 1; i <= 20; i++) {
    par[v][i] = par[par[v][i - 1]][i - 1];
  }
  for (auto to : adj[v]) {
    if (to == p) continue;
    dfs(to, v);
  }
  out_time[v] = tim++;
  upd(out_time[v], -a[v], 2 * n);
}
long long lca(long long u, long long v) {
  if (depth[u] > depth[v]) {
    swap(u, v);
  }
  long long diff = abs(depth[u] - depth[v]);
  for (long long i = 0; i <= 20; i++) {
    if (diff & (1ll << i)) {
      v = par[v][i];
    }
  }
  if (u == v) {
    return v;
  }
  for (long long i = 20; i >= 0; i--) {
    if (par[u][i] == par[v][i]) {
      continue;
    }
    u = par[u][i];
    v = par[v][i];
  }
  return par[u][0];
}
void solve(long long tc) {
  long long q, x, y;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  for (long long i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  long long ty;
  while (q--) {
    cin >> ty >> x >> y;
    if (ty == 1) {
      upd(in_time[x], -a[x], 2 * n);
      upd(out_time[x], a[x], 2 * n);
      a[x] = abs(y);
      upd(in_time[x], a[x], 2 * n);
      upd(out_time[x], -a[x], 2 * n);
    } else {
      long long lc = lca(x, y);
      long long d = 2 * (query(in_time[x]) + query(in_time[y]) -
                         2 * query(in_time[lc]) + a[lc]);
      cout << d - a[x] - a[y] << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  long long t = 1;
  long long tc = 1;
  while (t--) {
    solve(tc);
    tc++;
  }
  return 0;
}

```