#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
vector<int> adj[MAXN];
int sz[MAXN], dep[MAXN], top[MAXN], son[MAXN], rk[MAXN], dfn[MAXN], dfs_clock,
    fa[MAXN], n, q;
long long c[MAXN], sum[MAXN << 2];
void build(int rt, int l, int r) {
  if (l == r) {
    sum[rt] = c[rk[l]];
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void upd(int rt, int l, int r, int pos, long long x) {
  if (l == r) {
    sum[rt] = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) {
    upd(rt << 1, l, mid, pos, x);
  } else {
    upd(rt << 1 | 1, mid + 1, r, pos, x);
  }
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
long long qry(int rt, int l, int r, int L, int R) {
  if (L <= l && R >= r) {
    return sum[rt];
  }
  int mid = (l + r) >> 1;
  if (R <= mid) {
    return qry(rt << 1, l, mid, L, R);
  } else if (L > mid) {
    return qry(rt << 1 | 1, mid + 1, r, L, R);
  } else {
    return qry(rt << 1, l, mid, L, mid) +
           qry(rt << 1 | 1, mid + 1, r, mid + 1, R);
  }
}
void dfs1(int u, int d) {
  sz[u] = 1;
  son[u] = 0;
  dep[u] = d;
  for (int v : adj[u])
    if (fa[u] != v) {
      fa[v] = u;
      dfs1(v, d + 1);
      sz[u] += sz[v];
      if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;
    }
}
void dfs2(int u, int t) {
  top[u] = t;
  dfn[u] = ++dfs_clock;
  rk[dfs_clock] = u;
  if (son[u]) dfs2(son[u], t);
  for (int v : adj[u])
    if (son[u] != v && fa[u] != v) {
      dfs2(v, v);
    }
}
long long get(int u, int v) {
  long long res = 0;
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) {
      swap(u, v);
    }
    res += qry(1, 1, n, dfn[top[u]], dfn[u]);
    u = fa[top[u]];
  }
  if (dfn[u] > dfn[v]) swap(u, v);
  res += qry(1, 1, n, dfn[u], dfn[v]);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    c[i] = abs(c[i]);
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int op;
    cin >> op;
    if (op == 1) {
      int u;
      long long x;
      cin >> u >> x;
      x = abs(x);
      upd(1, 1, n, dfn[u], x);
      c[u] = x;
    } else {
      int u, v;
      cin >> u >> v;
      cout << 2ll * get(u, v) - c[u] - c[v] << '\n';
    }
  }
  return 0;
}

### Prompt

Generate 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 = 1e5 + 5;
vector<int> adj[MAXN];
int sz[MAXN], dep[MAXN], top[MAXN], son[MAXN], rk[MAXN], dfn[MAXN], dfs_clock,
    fa[MAXN], n, q;
long long c[MAXN], sum[MAXN << 2];
void build(int rt, int l, int r) {
  if (l == r) {
    sum[rt] = c[rk[l]];
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void upd(int rt, int l, int r, int pos, long long x) {
  if (l == r) {
    sum[rt] = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) {
    upd(rt << 1, l, mid, pos, x);
  } else {
    upd(rt << 1 | 1, mid + 1, r, pos, x);
  }
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
long long qry(int rt, int l, int r, int L, int R) {
  if (L <= l && R >= r) {
    return sum[rt];
  }
  int mid = (l + r) >> 1;
  if (R <= mid) {
    return qry(rt << 1, l, mid, L, R);
  } else if (L > mid) {
    return qry(rt << 1 | 1, mid + 1, r, L, R);
  } else {
    return qry(rt << 1, l, mid, L, mid) +
           qry(rt << 1 | 1, mid + 1, r, mid + 1, R);
  }
}
void dfs1(int u, int d) {
  sz[u] = 1;
  son[u] = 0;
  dep[u] = d;
  for (int v : adj[u])
    if (fa[u] != v) {
      fa[v] = u;
      dfs1(v, d + 1);
      sz[u] += sz[v];
      if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;
    }
}
void dfs2(int u, int t) {
  top[u] = t;
  dfn[u] = ++dfs_clock;
  rk[dfs_clock] = u;
  if (son[u]) dfs2(son[u], t);
  for (int v : adj[u])
    if (son[u] != v && fa[u] != v) {
      dfs2(v, v);
    }
}
long long get(int u, int v) {
  long long res = 0;
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) {
      swap(u, v);
    }
    res += qry(1, 1, n, dfn[top[u]], dfn[u]);
    u = fa[top[u]];
  }
  if (dfn[u] > dfn[v]) swap(u, v);
  res += qry(1, 1, n, dfn[u], dfn[v]);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    c[i] = abs(c[i]);
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int op;
    cin >> op;
    if (op == 1) {
      int u;
      long long x;
      cin >> u >> x;
      x = abs(x);
      upd(1, 1, n, dfn[u], x);
      c[u] = x;
    } else {
      int u, v;
      cin >> u >> v;
      cout << 2ll * get(u, v) - c[u] - c[v] << '\n';
    }
  }
  return 0;
}

```