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 dir[5] = {0, -1, 0, 1, 0};
const int maxn = 100010;
const int oo = 1e9;
const int mod = 1000000007;
int n, q, timeDFS, lg2, table[maxn][20], h[maxn];
long long a[maxn], dis[maxn], f[maxn];
pair<int, int> euler[maxn];
vector<int> graph[maxn];
void EUL(int u, int p, int d) {
  euler[u].first = ++timeDFS;
  dis[timeDFS] = d;
  for (int j = 1; j <= lg2; ++j) table[u][j] = table[table[u][j - 1]][j - 1];
  for (int v : graph[u])
    if (v != p) {
      table[v][0] = u;
      h[v] = h[u] + 1;
      EUL(v, u, d + a[v]);
    }
  euler[u].second = timeDFS;
}
void update(int p, long long k) {
  for (; p <= n; p += p & -p) f[p] += k;
}
long long get(int p) {
  long long res = 0;
  for (; p; p -= p & -p) res += f[p];
  return res;
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  for (int j = lg2; j >= 0; --j)
    if (h[table[u][j]] >= h[v]) u = table[u][j];
  if (u == v) return u;
  for (int j = lg2; j >= 0; --j)
    if (table[u][j] != table[v][j]) {
      u = table[u][j];
      v = table[v][j];
    }
  return table[u][0];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  long long u, v;
  for (int i = 1; i <= n - 1; ++i) {
    cin >> u >> v;
    graph[u].emplace_back(v);
    graph[v].emplace_back(u);
  }
  lg2 = log2(n);
  table[1][0] = 1;
  EUL(1, -1, a[1]);
  for (int i = 1; i <= n; ++i) {
    update(i, dis[i]);
    update(i + 1, -dis[i]);
  }
  int t;
  while (q--) {
    cin >> t >> u >> v;
    if (t == 1) {
      v = abs(v);
      update(euler[u].first, v - a[u]);
      update(euler[u].second + 1, a[u] - v);
      a[u] = v;
    } else {
      int anc = lca(u, v);
      long long du = get(euler[u].first);
      long long dv = get(euler[v].first);
      long long da = get(euler[anc].first);
      cout << du * 2 + dv * 2 - da * 4 - a[u] - a[v] + a[anc] * 2 << '\n';
    }
  }
}
