#include <bits/stdc++.h>
using namespace std;
template <typename T>
class BinaryIndexedTree {
 private:
  vector<T> c;

 public:
  void Build(int n) { c = vector<T>(n + 1, 0); }
  void Add(int x, T v) {
    for (; x < c.size(); x += x & -x) c[x] += v;
  }
  T Sum(int x) {
    T ret = 0;
    for (; x; x -= x & -x) ret += c[x];
    return ret;
  }
};
void solve() {
  int n, q;
  cin >> n >> q;
  vector<long long> a(n + 1);
  vector<vector<int>> G(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int N = 19;
  int dfs_clock = 0;
  vector<int> dep(n + 1), pre(n + 1), post(n + 1);
  vector<vector<int>> fa(n + 1, vector<int>(N + 1));
  function<void(int)> dfs = [&](int u) {
    for (int i = 1; i <= N; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
    pre[u] = ++dfs_clock;
    for (int v : G[u]) {
      if (v == fa[u][0]) continue;
      fa[v][0] = u;
      dep[v] = dep[u] + 1;
      dfs(v);
    }
    post[u] = dfs_clock;
  };
  dep[1] = 1;
  dfs(1);
  auto get_lca = [&](int u, int v) {
    if (dep[u] < dep[v]) swap(u, v);
    for (int i = N; i >= 0; i--)
      if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];
    for (int i = N; i >= 0; i--)
      if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
    if (u != v) u = fa[u][0];
    return u;
  };
  BinaryIndexedTree<long long> bit;
  bit.Build(n);
  auto modify = [&](int u, int c) {
    c = abs(c);
    bit.Add(pre[u], c - a[u]);
    bit.Add(post[u] + 1, a[u] - c);
    a[u] = c;
  };
  auto query = [&](int u, int v) {
    if (u == v) return 0LL;
    int lca = get_lca(u, v);
    long long ans = 0;
    ans += bit.Sum(pre[u]) + bit.Sum(pre[v]);
    ans -= bit.Sum(pre[lca]) + bit.Sum(pre[fa[lca][0]]);
    ans *= 2;
    ans -= a[u] + a[v];
    return ans;
  };
  for (int i = 1; i <= n; i++) {
    int w = a[i];
    a[i] = 0;
    modify(i, w);
  }
  for (int i = 0; i < q; i++) {
    int op, x, y;
    cin >> op >> x >> y;
    if (op == 1)
      modify(x, y);
    else
      cout << query(x, y) << '\n';
  }
}
int main() {
  cin.tie(0), cout.tie(0);
  ios::sync_with_stdio(false);
  solve();
}

### Prompt

Create a solution in CPP for 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;
template <typename T>
class BinaryIndexedTree {
 private:
  vector<T> c;

 public:
  void Build(int n) { c = vector<T>(n + 1, 0); }
  void Add(int x, T v) {
    for (; x < c.size(); x += x & -x) c[x] += v;
  }
  T Sum(int x) {
    T ret = 0;
    for (; x; x -= x & -x) ret += c[x];
    return ret;
  }
};
void solve() {
  int n, q;
  cin >> n >> q;
  vector<long long> a(n + 1);
  vector<vector<int>> G(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int N = 19;
  int dfs_clock = 0;
  vector<int> dep(n + 1), pre(n + 1), post(n + 1);
  vector<vector<int>> fa(n + 1, vector<int>(N + 1));
  function<void(int)> dfs = [&](int u) {
    for (int i = 1; i <= N; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
    pre[u] = ++dfs_clock;
    for (int v : G[u]) {
      if (v == fa[u][0]) continue;
      fa[v][0] = u;
      dep[v] = dep[u] + 1;
      dfs(v);
    }
    post[u] = dfs_clock;
  };
  dep[1] = 1;
  dfs(1);
  auto get_lca = [&](int u, int v) {
    if (dep[u] < dep[v]) swap(u, v);
    for (int i = N; i >= 0; i--)
      if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];
    for (int i = N; i >= 0; i--)
      if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
    if (u != v) u = fa[u][0];
    return u;
  };
  BinaryIndexedTree<long long> bit;
  bit.Build(n);
  auto modify = [&](int u, int c) {
    c = abs(c);
    bit.Add(pre[u], c - a[u]);
    bit.Add(post[u] + 1, a[u] - c);
    a[u] = c;
  };
  auto query = [&](int u, int v) {
    if (u == v) return 0LL;
    int lca = get_lca(u, v);
    long long ans = 0;
    ans += bit.Sum(pre[u]) + bit.Sum(pre[v]);
    ans -= bit.Sum(pre[lca]) + bit.Sum(pre[fa[lca][0]]);
    ans *= 2;
    ans -= a[u] + a[v];
    return ans;
  };
  for (int i = 1; i <= n; i++) {
    int w = a[i];
    a[i] = 0;
    modify(i, w);
  }
  for (int i = 0; i < q; i++) {
    int op, x, y;
    cin >> op >> x >> y;
    if (op == 1)
      modify(x, y);
    else
      cout << query(x, y) << '\n';
  }
}
int main() {
  cin.tie(0), cout.tie(0);
  ios::sync_with_stdio(false);
  solve();
}

```