#include <bits/stdc++.h>
using namespace std;
struct monoid {
  int L, R;
  long long sum;
  monoid() { L = 0, R = 0, sum = -1; }
  monoid(int x) { L = x, R = x, sum = 0; }
};
monoid merge(monoid A, monoid B) {
  if (A.sum == -1) {
    return B;
  }
  if (B.sum == -1) {
    return A;
  }
  monoid ans;
  ans.L = A.L;
  ans.R = B.R;
  ans.sum = A.sum + B.sum + max(abs(A.R + B.L), abs(A.R - B.L));
  return ans;
}
struct segment_tree {
  int N;
  array<vector<monoid>, 2> ST;
  segment_tree() {}
  segment_tree(vector<int> &A) {
    int N2 = A.size();
    N = 1;
    while (N < N2) {
      N *= 2;
    }
    ST[0] = vector<monoid>(N * 2 - 1);
    ST[1] = vector<monoid>(N * 2 - 1);
    for (int i = 0; i < N2; i++) {
      ST[0][N - 1 + i] = monoid(A[i]);
      ST[1][N - 1 + i] = monoid(A[i]);
    }
    for (int i = N - 2; i >= 0; i--) {
      ST[0][i] = merge(ST[0][i * 2 + 1], ST[0][i * 2 + 2]);
      ST[1][i] = merge(ST[1][i * 2 + 2], ST[1][i * 2 + 1]);
    }
  }
  void update(int i, int x) {
    i += N - 1;
    ST[0][i] = monoid(x);
    ST[1][i] = monoid(x);
    while (i > 0) {
      i = (i - 1) / 2;
      ST[0][i] = merge(ST[0][i * 2 + 1], ST[0][i * 2 + 2]);
      ST[1][i] = merge(ST[1][i * 2 + 2], ST[1][i * 2 + 1]);
    }
  }
  monoid fold(int L, int R, int d, int i, int l, int r) {
    if (r <= L || R <= l) {
      return monoid();
    } else if (L <= l && r <= R) {
      return ST[d][i];
    } else {
      int m = (l + r) / 2;
      if (d == 0) {
        return merge(fold(L, R, d, i * 2 + 1, l, m),
                     fold(L, R, d, i * 2 + 2, m, r));
      } else {
        return merge(fold(L, R, d, i * 2 + 2, m, r),
                     fold(L, R, d, i * 2 + 1, l, m));
      }
    }
  }
  monoid fold(int L, int R, int d) { return fold(L, R, d, 0, 0, N); }
};
struct heavy_light_decomposition {
  vector<int> p, sz, in, next;
  segment_tree ST;
  heavy_light_decomposition(vector<int> &p, vector<vector<int>> &c,
                            vector<int> &a)
      : p(p) {
    int N = p.size();
    sz = vector<int>(N);
    dfs1(c);
    in = vector<int>(N);
    next = vector<int>(N);
    next[0] = 0;
    int t = 0;
    dfs2(c, t);
    vector<int> a2(N);
    for (int i = 0; i < N; i++) {
      a2[in[i]] = a[i];
    }
    ST = segment_tree(a2);
  }
  void dfs1(vector<vector<int>> &c, int v = 0) {
    sz[v] = 1;
    for (int &w : c[v]) {
      dfs1(c, w);
      sz[v] += sz[w];
      if (sz[w] > sz[c[v][0]]) {
        swap(w, c[v][0]);
      }
    }
  }
  void dfs2(vector<vector<int>> &c, int &t, int v = 0) {
    in[v] = t;
    t++;
    for (int w : c[v]) {
      if (w == c[v][0]) {
        next[w] = next[v];
      } else {
        next[w] = w;
      }
      dfs2(c, t, w);
    }
  }
  int lca(int u, int v) {
    while (true) {
      if (in[u] > in[v]) {
        swap(u, v);
      }
      if (next[u] == next[v]) {
        return u;
      }
      v = p[next[v]];
    }
  }
  void update(int v, int x) { ST.update(in[v], x); }
  monoid query(int u, int v) {
    int w = lca(u, v);
    monoid ans1;
    while (next[u] != next[w]) {
      ans1 = merge(ans1, ST.fold(in[next[u]], in[u] + 1, 1));
      u = p[next[u]];
    }
    ans1 = merge(ans1, ST.fold(in[w], in[u] + 1, 1));
    monoid ans2;
    while (next[v] != next[w]) {
      ans2 = merge(ST.fold(in[next[v]], in[v] + 1, 0), ans2);
      v = p[next[v]];
    }
    ans2 = merge(ST.fold(in[w] + 1, in[v] + 1, 0), ans2);
    return merge(ans1, ans2);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<int>> E(n);
  for (int i = 0; i < n - 1; i++) {
    int s, t;
    cin >> s >> t;
    s--;
    t--;
    E[s].push_back(t);
    E[t].push_back(s);
  }
  vector<int> pr(n, -1);
  vector<vector<int>> ch(n);
  queue<int> Q;
  Q.push(0);
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    for (int w : E[v]) {
      if (w != pr[v]) {
        pr[w] = v;
        ch[v].push_back(w);
        Q.push(w);
      }
    }
  }
  heavy_light_decomposition T(pr, ch, a);
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int u, c;
      cin >> u >> c;
      u--;
      T.update(u, c);
    }
    if (t == 2) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      cout << T.query(u, v).sum << "\n";
    }
  }
}

### Prompt

Please formulate 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;
struct monoid {
  int L, R;
  long long sum;
  monoid() { L = 0, R = 0, sum = -1; }
  monoid(int x) { L = x, R = x, sum = 0; }
};
monoid merge(monoid A, monoid B) {
  if (A.sum == -1) {
    return B;
  }
  if (B.sum == -1) {
    return A;
  }
  monoid ans;
  ans.L = A.L;
  ans.R = B.R;
  ans.sum = A.sum + B.sum + max(abs(A.R + B.L), abs(A.R - B.L));
  return ans;
}
struct segment_tree {
  int N;
  array<vector<monoid>, 2> ST;
  segment_tree() {}
  segment_tree(vector<int> &A) {
    int N2 = A.size();
    N = 1;
    while (N < N2) {
      N *= 2;
    }
    ST[0] = vector<monoid>(N * 2 - 1);
    ST[1] = vector<monoid>(N * 2 - 1);
    for (int i = 0; i < N2; i++) {
      ST[0][N - 1 + i] = monoid(A[i]);
      ST[1][N - 1 + i] = monoid(A[i]);
    }
    for (int i = N - 2; i >= 0; i--) {
      ST[0][i] = merge(ST[0][i * 2 + 1], ST[0][i * 2 + 2]);
      ST[1][i] = merge(ST[1][i * 2 + 2], ST[1][i * 2 + 1]);
    }
  }
  void update(int i, int x) {
    i += N - 1;
    ST[0][i] = monoid(x);
    ST[1][i] = monoid(x);
    while (i > 0) {
      i = (i - 1) / 2;
      ST[0][i] = merge(ST[0][i * 2 + 1], ST[0][i * 2 + 2]);
      ST[1][i] = merge(ST[1][i * 2 + 2], ST[1][i * 2 + 1]);
    }
  }
  monoid fold(int L, int R, int d, int i, int l, int r) {
    if (r <= L || R <= l) {
      return monoid();
    } else if (L <= l && r <= R) {
      return ST[d][i];
    } else {
      int m = (l + r) / 2;
      if (d == 0) {
        return merge(fold(L, R, d, i * 2 + 1, l, m),
                     fold(L, R, d, i * 2 + 2, m, r));
      } else {
        return merge(fold(L, R, d, i * 2 + 2, m, r),
                     fold(L, R, d, i * 2 + 1, l, m));
      }
    }
  }
  monoid fold(int L, int R, int d) { return fold(L, R, d, 0, 0, N); }
};
struct heavy_light_decomposition {
  vector<int> p, sz, in, next;
  segment_tree ST;
  heavy_light_decomposition(vector<int> &p, vector<vector<int>> &c,
                            vector<int> &a)
      : p(p) {
    int N = p.size();
    sz = vector<int>(N);
    dfs1(c);
    in = vector<int>(N);
    next = vector<int>(N);
    next[0] = 0;
    int t = 0;
    dfs2(c, t);
    vector<int> a2(N);
    for (int i = 0; i < N; i++) {
      a2[in[i]] = a[i];
    }
    ST = segment_tree(a2);
  }
  void dfs1(vector<vector<int>> &c, int v = 0) {
    sz[v] = 1;
    for (int &w : c[v]) {
      dfs1(c, w);
      sz[v] += sz[w];
      if (sz[w] > sz[c[v][0]]) {
        swap(w, c[v][0]);
      }
    }
  }
  void dfs2(vector<vector<int>> &c, int &t, int v = 0) {
    in[v] = t;
    t++;
    for (int w : c[v]) {
      if (w == c[v][0]) {
        next[w] = next[v];
      } else {
        next[w] = w;
      }
      dfs2(c, t, w);
    }
  }
  int lca(int u, int v) {
    while (true) {
      if (in[u] > in[v]) {
        swap(u, v);
      }
      if (next[u] == next[v]) {
        return u;
      }
      v = p[next[v]];
    }
  }
  void update(int v, int x) { ST.update(in[v], x); }
  monoid query(int u, int v) {
    int w = lca(u, v);
    monoid ans1;
    while (next[u] != next[w]) {
      ans1 = merge(ans1, ST.fold(in[next[u]], in[u] + 1, 1));
      u = p[next[u]];
    }
    ans1 = merge(ans1, ST.fold(in[w], in[u] + 1, 1));
    monoid ans2;
    while (next[v] != next[w]) {
      ans2 = merge(ST.fold(in[next[v]], in[v] + 1, 0), ans2);
      v = p[next[v]];
    }
    ans2 = merge(ST.fold(in[w] + 1, in[v] + 1, 0), ans2);
    return merge(ans1, ans2);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<int>> E(n);
  for (int i = 0; i < n - 1; i++) {
    int s, t;
    cin >> s >> t;
    s--;
    t--;
    E[s].push_back(t);
    E[t].push_back(s);
  }
  vector<int> pr(n, -1);
  vector<vector<int>> ch(n);
  queue<int> Q;
  Q.push(0);
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    for (int w : E[v]) {
      if (w != pr[v]) {
        pr[w] = v;
        ch[v].push_back(w);
        Q.push(w);
      }
    }
  }
  heavy_light_decomposition T(pr, ch, a);
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int u, c;
      cin >> u >> c;
      u--;
      T.update(u, c);
    }
    if (t == 2) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      cout << T.query(u, v).sum << "\n";
    }
  }
}

```