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;
#define ll long long

const int mx = 112345;
vector<int> adj[mx];
int h[mx], sz[mx], par[mx][20], in[mx], t = 0;


int bit[mx];
inline int lb (int x) { return x & -x; };

ll qry (int x) {
  ll ans = 0;
  while(x) ans += bit[x], x -= lb(x);
  return ans;
}

void upd (int x, int val) {
  ll ans = 0;
  while(x < mx) bit[x] += val, x += lb(x);
}

void dfs (int v) {
  in[v] = ++t; sz[v] = 1;
  for (auto u : adj[v]) {
    if (u == par[v][0]) continue;
    h[u] = 1 + h[v];
    par[u][0] = v;
    for (int i = 1; i < 20; i++) {
      par[u][i] = par[par[u][i-1]][i-1];
    }
    dfs(u); sz[v] += sz[u];
  }
}

int lca (int a, int b) {
  if (h[a] < h[b]) swap(a, b);
  for (int i = 19; i >= 0; i--) {
    if (h[par[a][i]] >= h[b]) a = par[a][i];
  }
  
  for (int i = 19; i >= 0; i--) {
    if (par[a][i] != par[b][i]) a = par[a][i], b = par[b][i];
  }
  if (a != b) a = par[a][0];
  return a;
}

int v[mx];
main () {
  ios_base::sync_with_stdio(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    v[i] = abs(v[i]);
  }
  
  for (int i = 0; i < n-1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  
  h[1] = 1, dfs(1);
  
  for (int i = 1; i <= n; i++) {
    upd (in[i], v[i]);
    upd (in[i]+sz[i], -v[i]);
  }
  
  for (int i = 0; i < q; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    if (a == 1) {
      c = abs(c)-v[i];
      upd (in[b], c);
      upd (in[b]+sz[i], -c);
      v[i] += c;
    } else {
      if (b == c) {
        cout << 0 << endl;
        continue;
      }
      int p = lca(b, c);
      ll nodeSum = qry(in[b]) + qry(in[c]) - qry(in[p]) - qry(in[par[p][0]]);
      cout << 2 * nodeSum - v[b] - v[c] << endl;
    }
  }
  
}
