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.
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, q, a[N];
vector<int> adj[N];
namespace cd {
int sz[N];
int cpar[N];
bool vis[N];
void dfs(int n, int p = -1) {
  sz[n] = 1;
  for (int v : adj[n])
    if (v != p && !vis[v]) {
      dfs(v, n);
      sz[n] += sz[v];
    }
}
int centroid(int n) {
  dfs(n);
  int num = sz[n];
  int p = -1;
  do {
    int nxt = -1;
    for (int v : adj[n])
      if (v != p && !vis[v]) {
        if (2 * sz[v] > num) nxt = v;
      }
    p = n, n = nxt;
  } while (~n);
  return p;
}
void centroid_decomp(int n = 0, int p = -1) {
  int c = centroid(n);
  vis[c] = true;
  cpar[c] = p;
  for (int v : adj[c])
    if (!vis[v]) {
      centroid_decomp(v, c);
    }
}
}  // namespace cd
using cd::centroid_decomp;
using cd::cpar;
vector<vector<int> > treeJump(vector<int>& P) {
  int on = 1, d = 1;
  while (on < (int)(P).size()) on *= 2, d++;
  vector<vector<int> > jmp(d, P);
  for (int i = 1; i < (d); ++i)
    for (int j = 0; j < ((int)(P).size()); ++j)
      jmp[i][j] = jmp[i - 1][jmp[i - 1][j]];
  return jmp;
}
int jmp(vector<vector<int> >& tbl, int nod, int steps) {
  for (int i = 0; i < ((int)(tbl).size()); ++i)
    if (steps & (1 << i)) nod = tbl[i][nod];
  return nod;
}
int lca(vector<vector<int> >& tbl, vector<int>& depth, int a, int b) {
  if (depth[a] < depth[b]) swap(a, b);
  a = jmp(tbl, a, depth[a] - depth[b]);
  if (a == b) return a;
  for (int i = (int)(tbl).size(); i--;) {
    int c = tbl[i][a], d = tbl[i][b];
    if (c != d) a = c, b = d;
  }
  return tbl[0][a];
}
void dfs(int v, vector<int>& depth) {
  for (const int x : adj[v]) {
    depth[x] = depth[v] + 1;
    dfs(x, depth);
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  cin >> n >> q;
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  for (int i = 0; i < (n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  centroid_decomp();
  int root = -1;
  for (int i = 0; i < (n); ++i) {
    if (cpar[i] == -1) {
      cpar[i] = i;
      root = i;
    }
    adj[i].clear();
  }
  for (int i = 0; i < (n); ++i) {
    if (cpar[i] != i) {
      adj[cpar[i]].push_back(i);
    }
  }
  vector<int> P(cpar, cpar + n), depth(n);
  vector<vector<int> > tbl = treeJump(P);
  dfs(root, depth);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int u, c;
      cin >> u >> c;
      --u;
      a[u] = abs(c);
    } else {
      int u, v;
      cin >> u >> v;
      --u, --v;
      int anc = lca(tbl, depth, u, v);
      long long ans = 0;
      while (u != anc) {
        ans += a[u] + a[cpar[u]];
        u = cpar[u];
      }
      while (v != anc) {
        ans += a[v] + a[cpar[v]];
        v = cpar[v];
      }
      cout << ans << '\n';
    }
  }
}
