#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 9;
struct BIT {
  int n;
  long long s[N];
  void init(int _n) { n = _n; }
  void upd(int p, long long k) {
    for (; p <= n; p += p & -p) s[p] += k;
  }
  long long sum(int p) {
    long long r = 0;
    for (; p; p -= p & -p) r += s[p];
    return r;
  }
  long long sum(int l, int r) { return sum(r) - sum(l - 1); }
} bit;
long long a[N], b[N];
int tot, p[N], dep[N], sz[N], hc[N], top[N], dfn[N];
vector<int> G[N];
void dfs1(int u, int _p) {
  dep[u] = dep[p[u] = _p] + 1;
  sz[u] = 1;
  hc[u] = 0;
  for (int v : G[u])
    if (v != _p) {
      dfs1(v, u);
      sz[u] += sz[v];
      if (sz[v] > sz[hc[u]]) hc[u] = v;
    }
}
void dfs2(int u, int _top) {
  top[u] = _top;
  dfn[u] = ++tot;
  if (hc[u]) dfs2(hc[u], _top);
  for (int v : G[u])
    if (v != p[u] && v != hc[u]) dfs2(v, v);
}
void upd(int u, int c) {
  a[u] = c;
  if (u != 1) {
    long long pre = b[dfn[u]];
    b[dfn[u]] = max(abs(a[u] + a[p[u]]), abs(a[u] - a[p[u]]));
    bit.upd(dfn[u], b[dfn[u]] - pre);
  }
  if (hc[u]) {
    int v = hc[u];
    long long pre = b[dfn[v]];
    b[dfn[v]] = max(abs(a[v] + a[p[v]]), abs(a[v] - a[p[v]]));
    bit.upd(dfn[v], b[dfn[v]] - pre);
  }
}
long long query(int u, int v) {
  long long dis = 0;
  for (; top[u] != top[v]; u = p[top[u]]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    dis += bit.sum(dfn[top[u]], dfn[u]);
    int w = top[u];
    dis += max(abs(a[w] + a[p[w]]), abs(a[w] - a[p[w]])) - b[dfn[w]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  if (u != v) dis += bit.sum(dfn[u] + 1, dfn[v]);
  return dis;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  bit.init(n);
  for (int i = 2; i <= n; ++i) {
    b[dfn[i]] = max(abs(a[i] + a[p[i]]), abs(a[i] - a[p[i]]));
    bit.upd(dfn[i], b[dfn[i]]);
  }
  for (int i = 1; i <= q; ++i) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int u, c;
      scanf("%d %d", &u, &c);
      upd(u, c);
    } else if (op == 2) {
      int u, v;
      scanf("%d %d", &u, &v);
      long long ans = query(u, v);
      printf("%lld\n", ans);
    }
  }
  return 0;
}

### 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;
const int N = 100000 + 9;
struct BIT {
  int n;
  long long s[N];
  void init(int _n) { n = _n; }
  void upd(int p, long long k) {
    for (; p <= n; p += p & -p) s[p] += k;
  }
  long long sum(int p) {
    long long r = 0;
    for (; p; p -= p & -p) r += s[p];
    return r;
  }
  long long sum(int l, int r) { return sum(r) - sum(l - 1); }
} bit;
long long a[N], b[N];
int tot, p[N], dep[N], sz[N], hc[N], top[N], dfn[N];
vector<int> G[N];
void dfs1(int u, int _p) {
  dep[u] = dep[p[u] = _p] + 1;
  sz[u] = 1;
  hc[u] = 0;
  for (int v : G[u])
    if (v != _p) {
      dfs1(v, u);
      sz[u] += sz[v];
      if (sz[v] > sz[hc[u]]) hc[u] = v;
    }
}
void dfs2(int u, int _top) {
  top[u] = _top;
  dfn[u] = ++tot;
  if (hc[u]) dfs2(hc[u], _top);
  for (int v : G[u])
    if (v != p[u] && v != hc[u]) dfs2(v, v);
}
void upd(int u, int c) {
  a[u] = c;
  if (u != 1) {
    long long pre = b[dfn[u]];
    b[dfn[u]] = max(abs(a[u] + a[p[u]]), abs(a[u] - a[p[u]]));
    bit.upd(dfn[u], b[dfn[u]] - pre);
  }
  if (hc[u]) {
    int v = hc[u];
    long long pre = b[dfn[v]];
    b[dfn[v]] = max(abs(a[v] + a[p[v]]), abs(a[v] - a[p[v]]));
    bit.upd(dfn[v], b[dfn[v]] - pre);
  }
}
long long query(int u, int v) {
  long long dis = 0;
  for (; top[u] != top[v]; u = p[top[u]]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    dis += bit.sum(dfn[top[u]], dfn[u]);
    int w = top[u];
    dis += max(abs(a[w] + a[p[w]]), abs(a[w] - a[p[w]])) - b[dfn[w]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  if (u != v) dis += bit.sum(dfn[u] + 1, dfn[v]);
  return dis;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  bit.init(n);
  for (int i = 2; i <= n; ++i) {
    b[dfn[i]] = max(abs(a[i] + a[p[i]]), abs(a[i] - a[p[i]]));
    bit.upd(dfn[i], b[dfn[i]]);
  }
  for (int i = 1; i <= q; ++i) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int u, c;
      scanf("%d %d", &u, &c);
      upd(u, c);
    } else if (op == 2) {
      int u, v;
      scanf("%d %d", &u, &v);
      long long ans = query(u, v);
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```