#include <bits/stdc++.h>
using namespace std;
int n, q;
int a[100005];
int de[100005];
int pr[100005][20];
int sa[100005], so[100005];
int cl[100005];
long long b[100005];
int id = 0;
vector<int> g[100005];
void update(int u, int delta) {
  while (u <= n) b[u] += delta, u += u & (-u);
}
long long get(int u) {
  long long res = 0;
  while (u > 0) {
    res += b[u];
    u -= u & (-u);
  }
  return res;
}
void DFS(int u) {
  cl[u] = 1;
  sa[u] = ++id;
  for (auto &v : g[u])
    if (!cl[v]) pr[v][0] = u, de[v] = de[u] + 1, DFS(v);
  so[u] = id;
}
int LCA(int u, int v) {
  while (de[u] > de[v]) {
    int h = log2(de[u] - de[v]);
    u = pr[u][h];
  }
  while (de[v] > de[u]) {
    int h = log2(de[v] - de[u]);
    v = pr[v][h];
  }
  if (u == v) return u;
  int h = log2(de[u]);
  while (h >= 0) {
    if (pr[u][h] != pr[v][h]) u = pr[u][h], v = pr[v][h];
    h--;
  }
  u = pr[u][0];
  return u;
}
int main() {
  ios ::sync_with_stdio(false);
  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);
  }
  de[1] = 1;
  DFS(1);
  int k = log2(n);
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= n; ++j) pr[j][i] = pr[pr[j][i - 1]][i - 1];
  for (int i = 1; i <= n; ++i) update(sa[i], a[i]), update(so[i] + 1, -a[i]);
  for (int i = 1; i <= q; ++i) {
    int tp, u, v;
    cin >> tp >> u >> v;
    if (tp == 1) {
      update(sa[u], -a[u]), update(so[u] + 1, a[u]);
      a[u] = abs(v);
      update(sa[u], a[u]), update(so[u] + 1, -a[u]);
    } else {
      int z = LCA(u, v);
      long long ans = get(sa[u]) + get(sa[v]);
      ans -= 2 * get(sa[z]);
      ans *= 2;
      ans += 2 * a[z];
      ans -= a[u] + a[v];
      cout << ans << "\n";
    }
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int n, q;
int a[100005];
int de[100005];
int pr[100005][20];
int sa[100005], so[100005];
int cl[100005];
long long b[100005];
int id = 0;
vector<int> g[100005];
void update(int u, int delta) {
  while (u <= n) b[u] += delta, u += u & (-u);
}
long long get(int u) {
  long long res = 0;
  while (u > 0) {
    res += b[u];
    u -= u & (-u);
  }
  return res;
}
void DFS(int u) {
  cl[u] = 1;
  sa[u] = ++id;
  for (auto &v : g[u])
    if (!cl[v]) pr[v][0] = u, de[v] = de[u] + 1, DFS(v);
  so[u] = id;
}
int LCA(int u, int v) {
  while (de[u] > de[v]) {
    int h = log2(de[u] - de[v]);
    u = pr[u][h];
  }
  while (de[v] > de[u]) {
    int h = log2(de[v] - de[u]);
    v = pr[v][h];
  }
  if (u == v) return u;
  int h = log2(de[u]);
  while (h >= 0) {
    if (pr[u][h] != pr[v][h]) u = pr[u][h], v = pr[v][h];
    h--;
  }
  u = pr[u][0];
  return u;
}
int main() {
  ios ::sync_with_stdio(false);
  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);
  }
  de[1] = 1;
  DFS(1);
  int k = log2(n);
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= n; ++j) pr[j][i] = pr[pr[j][i - 1]][i - 1];
  for (int i = 1; i <= n; ++i) update(sa[i], a[i]), update(so[i] + 1, -a[i]);
  for (int i = 1; i <= q; ++i) {
    int tp, u, v;
    cin >> tp >> u >> v;
    if (tp == 1) {
      update(sa[u], -a[u]), update(so[u] + 1, a[u]);
      a[u] = abs(v);
      update(sa[u], a[u]), update(so[u] + 1, -a[u]);
    } else {
      int z = LCA(u, v);
      long long ans = get(sa[u]) + get(sa[v]);
      ans -= 2 * get(sa[z]);
      ans *= 2;
      ans += 2 * a[z];
      ans -= a[u] + a[v];
      cout << ans << "\n";
    }
  }
}

```