#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
vector<int> g[maxn];
int f[maxn][20], dep[maxn], a[maxn], id[maxn], sz[maxn], dcnt = 0;
void dfs(int u, int fa) {
  id[u] = ++dcnt;
  sz[u] = 1;
  for (int v : g[u]) {
    if (v == fa) continue;
    dep[v] = dep[u] + 1;
    f[v][0] = u;
    for (int i = 1; i <= 19; ++i) f[v][i] = f[f[v][i - 1]][i - 1];
    dfs(v, u);
    sz[u] += sz[v];
  }
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int d = dep[u] - dep[v], i = 19; i >= 0; --i)
    if (d & (1 << i)) u = f[u][i];
  if (u == v) return u;
  for (int i = 19; i >= 0; --i)
    if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];
  return f[u][0];
}
long long s[maxn << 2], lz[maxn << 2];
void update(int L, int R, int v, int l, int r, int rt) {
  if (L <= l && r <= R) {
    s[rt] += 1ll * v * (r - l + 1);
    lz[rt] += v;
    return;
  }
  int m = (l + r) / 2;
  if (lz[rt]) {
    lz[rt << 1] += lz[rt];
    lz[rt << 1 | 1] += lz[rt];
    s[rt << 1] += lz[rt] * (m - l + 1);
    s[rt << 1 | 1] += lz[rt] * (r - m);
    lz[rt] = 0;
  }
  if (L <= m) update(L, R, v, l, m, rt << 1);
  if (m < R) update(L, R, v, m + 1, r, rt << 1 | 1);
  s[rt] = s[rt << 1] + s[rt << 1 | 1];
}
long long query(int p, int l, int r, int rt) {
  if (l == r) return s[rt];
  int m = (l + r) / 2;
  if (lz[rt]) {
    lz[rt << 1] += lz[rt];
    lz[rt << 1 | 1] += lz[rt];
    s[rt << 1] += lz[rt] * (m - l + 1);
    s[rt << 1 | 1] += lz[rt] * (r - m);
    lz[rt] = 0;
  }
  if (p <= m) return query(p, l, m, rt << 1);
  return query(p, m + 1, r, rt << 1 | 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i], a[i] = abs(a[i]);
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) update(id[i], id[i] + sz[i] - 1, a[i], 1, n, 1);
  while (q--) {
    int op, u, v;
    cin >> op >> u >> v;
    if (op == 1) {
      update(id[u], id[u] + sz[u] - 1, -a[u], 1, n, 1);
      a[u] = abs(v);
      update(id[u], id[u] + sz[u] - 1, a[u], 1, n, 1);
    } else {
      int p = lca(u, v);
      long long ans = query(id[u], 1, n, 1) + query(id[v], 1, n, 1);
      if (p != 1) ans -= 2 * query(id[f[p][0]], 1, n, 1);
      ans -= a[p];
      ans *= 2;
      ans -= a[u], ans -= a[v];
      cout << ans << '\n';
    }
  }
}

### Prompt

Your task is to create a CPP solution to 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;
const int maxn = 100005;
vector<int> g[maxn];
int f[maxn][20], dep[maxn], a[maxn], id[maxn], sz[maxn], dcnt = 0;
void dfs(int u, int fa) {
  id[u] = ++dcnt;
  sz[u] = 1;
  for (int v : g[u]) {
    if (v == fa) continue;
    dep[v] = dep[u] + 1;
    f[v][0] = u;
    for (int i = 1; i <= 19; ++i) f[v][i] = f[f[v][i - 1]][i - 1];
    dfs(v, u);
    sz[u] += sz[v];
  }
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int d = dep[u] - dep[v], i = 19; i >= 0; --i)
    if (d & (1 << i)) u = f[u][i];
  if (u == v) return u;
  for (int i = 19; i >= 0; --i)
    if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];
  return f[u][0];
}
long long s[maxn << 2], lz[maxn << 2];
void update(int L, int R, int v, int l, int r, int rt) {
  if (L <= l && r <= R) {
    s[rt] += 1ll * v * (r - l + 1);
    lz[rt] += v;
    return;
  }
  int m = (l + r) / 2;
  if (lz[rt]) {
    lz[rt << 1] += lz[rt];
    lz[rt << 1 | 1] += lz[rt];
    s[rt << 1] += lz[rt] * (m - l + 1);
    s[rt << 1 | 1] += lz[rt] * (r - m);
    lz[rt] = 0;
  }
  if (L <= m) update(L, R, v, l, m, rt << 1);
  if (m < R) update(L, R, v, m + 1, r, rt << 1 | 1);
  s[rt] = s[rt << 1] + s[rt << 1 | 1];
}
long long query(int p, int l, int r, int rt) {
  if (l == r) return s[rt];
  int m = (l + r) / 2;
  if (lz[rt]) {
    lz[rt << 1] += lz[rt];
    lz[rt << 1 | 1] += lz[rt];
    s[rt << 1] += lz[rt] * (m - l + 1);
    s[rt << 1 | 1] += lz[rt] * (r - m);
    lz[rt] = 0;
  }
  if (p <= m) return query(p, l, m, rt << 1);
  return query(p, m + 1, r, rt << 1 | 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i], a[i] = abs(a[i]);
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) update(id[i], id[i] + sz[i] - 1, a[i], 1, n, 1);
  while (q--) {
    int op, u, v;
    cin >> op >> u >> v;
    if (op == 1) {
      update(id[u], id[u] + sz[u] - 1, -a[u], 1, n, 1);
      a[u] = abs(v);
      update(id[u], id[u] + sz[u] - 1, a[u], 1, n, 1);
    } else {
      int p = lca(u, v);
      long long ans = query(id[u], 1, n, 1) + query(id[v], 1, n, 1);
      if (p != 1) ans -= 2 * query(id[f[p][0]], 1, n, 1);
      ans -= a[p];
      ans *= 2;
      ans -= a[u], ans -= a[v];
      cout << ans << '\n';
    }
  }
}

```