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 maxn = 1e5 + 10;
int n, q, tot;
int a[maxn], id[maxn], getid[maxn];
int sz[maxn], fa[maxn], son[maxn], top[maxn], dep[maxn];
long long sum[maxn << 2];
vector<int> v[maxn];
void dfs1(int u, int f, int d) {
  dep[u] = d;
  fa[u] = f;
  sz[u] = 1;
  for (int i = 0; i < v[u].size(); i++) {
    if (v[u][i] == fa[u]) continue;
    dfs1(v[u][i], u, d + 1);
    sz[u] += sz[v[u][i]];
    if (sz[son[u]] < sz[v[u][i]]) son[u] = v[u][i];
  }
}
void dfs2(int u, int t) {
  top[u] = t;
  id[u] = ++tot;
  getid[tot] = u;
  if (son[u]) dfs2(son[u], t);
  for (int i = 0; i < v[u].size(); i++) {
    if (v[u][i] == fa[u] || v[u][i] == son[u]) continue;
    dfs2(v[u][i], v[u][i]);
  }
}
void pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
void build(int l, int r, int rt) {
  if (l == r) {
    sum[rt] = a[getid[l]];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1), build(mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
void mod(int l, int r, int rt, int pos, int v) {
  if (l == r) {
    sum[rt] = v;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) mod(l, mid, rt << 1, pos, v);
  if (mid < pos) mod(mid + 1, r, rt << 1 | 1, pos, v);
  pushup(rt);
}
long long que(int l, int r, int rt, int ql, int qr) {
  if (ql <= l && r <= qr) return sum[rt];
  int mid = (l + r) >> 1;
  long long ret = 0;
  if (ql <= mid) ret += que(l, mid, rt << 1, ql, qr);
  if (mid < qr) ret += que(mid + 1, r, rt << 1 | 1, ql, qr);
  return ret;
}
long long chainque(int x, int y) {
  long long ret = 0;
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    ret += que(1, n, 1, id[top[x]], id[x]);
    x = fa[top[x]];
  }
  if (dep[x] < dep[y]) swap(x, y);
  return ret + que(1, n, 1, id[y], id[x]);
}
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);
  }
  dfs1(1, 0, 0);
  dfs2(1, 1);
  build(1, n, 1);
  for (int i = 1, x, y, z; i <= q; i++) {
    scanf("%d%d%d", &x, &y, &z);
    if (x == 1) {
      z = abs(z);
      mod(1, n, 1, y, z);
      a[y] = z;
    } else if (x == 2) {
      printf("%lld\n", 2 * chainque(y, z) - a[y] - a[z]);
    }
  }
}
