#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
long long tree[MAXN];
vector<int> g[MAXN];
int curtime, ltime[MAXN], rtime[MAXN], par[20][MAXN], depth[MAXN];
void dfs(int u, int parent = -1) {
  par[0][u] = parent;
  for (int j = 1; j < 20; j++) {
    if (par[j - 1][u] == -1)
      par[j][u] = -1;
    else
      par[j][u] = par[j - 1][par[j - 1][u]];
  }
  ltime[u] = curtime++;
  rtime[u] = ltime[u];
  for (int v : g[u]) {
    if (v == parent) continue;
    depth[v] = depth[u] + 1;
    dfs(v, u);
    rtime[u] = rtime[v];
  }
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  int d = depth[u] - depth[v];
  for (int j = 0; j < 20; j++)
    if (d & (1 << j)) {
      u = par[j][u];
    }
  if (u == v) return u;
  for (int j = 19; j >= 0; j--)
    if (par[j][u] != par[j][v]) {
      u = par[j][u];
      v = par[j][v];
    }
  return par[0][u];
}
void add(int u, long long val) {
  u++;
  while (u < MAXN) {
    tree[u] += val;
    u += u & (-u);
  }
}
long long sum(int u) {
  u++;
  long long ret = 0;
  while (u > 0) {
    ret += tree[u];
    u -= u & (-u);
  }
  return ret;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i] = abs(a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++) {
    add(ltime[i], a[i]);
    add(rtime[i] + 1, -a[i]);
  }
  auto calc = [&](int u, int v) {
    long long cur = sum(ltime[u]) + sum(ltime[v]);
    int l = lca(u, v);
    cur -= sum(ltime[l]);
    if (par[0][l] != -1) cur -= sum(ltime[par[0][l]]);
    return cur;
  };
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int u, c;
      scanf("%d %d", &u, &c);
      u--;
      c = abs(c);
      add(ltime[u], c - a[u]);
      add(rtime[u] + 1, a[u] - c);
      a[u] = c;
    } else {
      int u, v;
      scanf("%d %d", &u, &v);
      u--, v--;
      if (u == v) {
        printf("0\n");
        continue;
      }
      long long cur = 2LL * calc(u, v);
      cur -= sum(ltime[u]);
      if (par[0][u] != -1) cur += sum(ltime[par[0][u]]);
      cur -= sum(ltime[v]);
      if (par[0][v] != -1) cur += sum(ltime[par[0][v]]);
      printf("%lld\n", cur);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long tree[MAXN];
vector<int> g[MAXN];
int curtime, ltime[MAXN], rtime[MAXN], par[20][MAXN], depth[MAXN];
void dfs(int u, int parent = -1) {
  par[0][u] = parent;
  for (int j = 1; j < 20; j++) {
    if (par[j - 1][u] == -1)
      par[j][u] = -1;
    else
      par[j][u] = par[j - 1][par[j - 1][u]];
  }
  ltime[u] = curtime++;
  rtime[u] = ltime[u];
  for (int v : g[u]) {
    if (v == parent) continue;
    depth[v] = depth[u] + 1;
    dfs(v, u);
    rtime[u] = rtime[v];
  }
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  int d = depth[u] - depth[v];
  for (int j = 0; j < 20; j++)
    if (d & (1 << j)) {
      u = par[j][u];
    }
  if (u == v) return u;
  for (int j = 19; j >= 0; j--)
    if (par[j][u] != par[j][v]) {
      u = par[j][u];
      v = par[j][v];
    }
  return par[0][u];
}
void add(int u, long long val) {
  u++;
  while (u < MAXN) {
    tree[u] += val;
    u += u & (-u);
  }
}
long long sum(int u) {
  u++;
  long long ret = 0;
  while (u > 0) {
    ret += tree[u];
    u -= u & (-u);
  }
  return ret;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i] = abs(a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++) {
    add(ltime[i], a[i]);
    add(rtime[i] + 1, -a[i]);
  }
  auto calc = [&](int u, int v) {
    long long cur = sum(ltime[u]) + sum(ltime[v]);
    int l = lca(u, v);
    cur -= sum(ltime[l]);
    if (par[0][l] != -1) cur -= sum(ltime[par[0][l]]);
    return cur;
  };
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int u, c;
      scanf("%d %d", &u, &c);
      u--;
      c = abs(c);
      add(ltime[u], c - a[u]);
      add(rtime[u] + 1, a[u] - c);
      a[u] = c;
    } else {
      int u, v;
      scanf("%d %d", &u, &v);
      u--, v--;
      if (u == v) {
        printf("0\n");
        continue;
      }
      long long cur = 2LL * calc(u, v);
      cur -= sum(ltime[u]);
      if (par[0][u] != -1) cur += sum(ltime[par[0][u]]);
      cur -= sum(ltime[v]);
      if (par[0][v] != -1) cur += sum(ltime[par[0][v]]);
      printf("%lld\n", cur);
    }
  }
  return 0;
}

```