#include <bits/stdc++.h>
using namespace std;
long double PI = acos(-1);
long double DEL = 1e-12;
long long M = 1e9 + 7;
const long long N = 3e5 + 10;
long long cnt, sum, mx, mn, ans, a[N], b[N];
long long n, m, d, i, j, k, l, p, q, r, t, w, x, y, z, tc;
string s;
struct HEAVY_LIGHT {
  vector<long long> dfsarray, depth, heavy, par, sub, head, pos, seg;
  void buildseg(long long i, long long l, long long r) {
    if (l == r) {
      seg[i] = a[dfsarray[l]];
      return;
    }
    buildseg(2 * i, l, ((l + r) >> 1));
    buildseg(2 * i + 1, ((l + r) >> 1) + 1, r);
    seg[i] = seg[2 * i] + seg[2 * i + 1];
  }
  void upd(long long x, long long v, long long i = 1, long long l = 0,
           long long r = n - 1) {
    if (x < l or x > r) {
      return;
    }
    if (l == r) {
      seg[i] = v;
      return;
    }
    upd(x, v, 2 * i, l, ((l + r) >> 1));
    upd(x, v, 2 * i + 1, ((l + r) >> 1) + 1, r);
    seg[i] = seg[2 * i] + seg[2 * i + 1];
  }
  long long query(long long a, long long b, long long i = 1, long long l = 0,
                  long long r = n - 1) {
    if (l > b or r < a) {
      return 0;
    }
    if (a <= l and r <= b) {
      return seg[i];
    }
    return query(a, b, 2 * i, l, ((l + r) >> 1)) +
           query(a, b, 2 * i + 1, ((l + r) >> 1) + 1, r);
  }
  void build(vector<vector<long long>> &g, long long root) {
    long long sz = n;
    depth.resize(sz + 1);
    heavy.resize(sz + 1);
    par.resize(sz + 1);
    pos.resize(sz + 1);
    head.resize(sz + 1);
    sub.resize(sz + 1);
    seg.resize(4 * sz + 1);
    dfs(root, root, g);
    decompose(root, root, g);
    buildseg(1, 0, n - 1);
  }
  void dfs(long long i, long long p, vector<vector<long long>> &g) {
    depth[i] = depth[p] + 1;
    par[i] = p;
    sub[i] = 1;
    long long ms = 0, hc = -1;
    for (auto j : g[i]) {
      if (j == p) continue;
      dfs(j, i, g);
      sub[i] += sub[j];
      if (sub[j] > ms) {
        ms = sub[j];
        hc = j;
      }
    }
    heavy[i] = hc;
  }
  void decompose(long long i, long long h, vector<vector<long long>> &g) {
    pos[i] = dfsarray.size();
    dfsarray.push_back(i);
    head[i] = h;
    if (heavy[i] != -1) {
      decompose(heavy[i], h, g);
    }
    for (auto j : g[i]) {
      if (j == par[i] or j == heavy[i]) continue;
      decompose(j, j, g);
    }
  }
  long long lca(long long u, long long v) {
    for (; head[u] != head[v]; u = par[head[u]]) {
      if (depth[head[u]] < depth[head[v]]) {
        swap(u, v);
      }
    }
    return depth[u] > depth[v] ? v : u;
  }
  long long get(long long u, long long v) {
    long long ans = 0;
    long long sub = a[u] + a[v];
    for (; head[u] != head[v]; u = par[head[u]]) {
      if (depth[head[u]] < depth[head[v]]) {
        swap(u, v);
      }
      ans += query(pos[head[u]], pos[u]);
    }
    if (depth[u] > depth[v]) {
      swap(u, v);
    }
    ans += query(pos[u], pos[v]);
    ans *= 2;
    return ans - sub;
  }
};
vector<vector<long long>> g;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  g.resize(n + 1);
  for (long long i = 1; i < n; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  HEAVY_LIGHT h;
  h.build(g, 1);
  while (q--) {
    cin >> t >> x >> y;
    if (t == 1) {
      h.upd(h.pos[x], abs(y));
      a[x] = abs(y);
    } else {
      cout << h.get(x, y);
      cout << "\n";
    }
  }
}

### 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;
long double PI = acos(-1);
long double DEL = 1e-12;
long long M = 1e9 + 7;
const long long N = 3e5 + 10;
long long cnt, sum, mx, mn, ans, a[N], b[N];
long long n, m, d, i, j, k, l, p, q, r, t, w, x, y, z, tc;
string s;
struct HEAVY_LIGHT {
  vector<long long> dfsarray, depth, heavy, par, sub, head, pos, seg;
  void buildseg(long long i, long long l, long long r) {
    if (l == r) {
      seg[i] = a[dfsarray[l]];
      return;
    }
    buildseg(2 * i, l, ((l + r) >> 1));
    buildseg(2 * i + 1, ((l + r) >> 1) + 1, r);
    seg[i] = seg[2 * i] + seg[2 * i + 1];
  }
  void upd(long long x, long long v, long long i = 1, long long l = 0,
           long long r = n - 1) {
    if (x < l or x > r) {
      return;
    }
    if (l == r) {
      seg[i] = v;
      return;
    }
    upd(x, v, 2 * i, l, ((l + r) >> 1));
    upd(x, v, 2 * i + 1, ((l + r) >> 1) + 1, r);
    seg[i] = seg[2 * i] + seg[2 * i + 1];
  }
  long long query(long long a, long long b, long long i = 1, long long l = 0,
                  long long r = n - 1) {
    if (l > b or r < a) {
      return 0;
    }
    if (a <= l and r <= b) {
      return seg[i];
    }
    return query(a, b, 2 * i, l, ((l + r) >> 1)) +
           query(a, b, 2 * i + 1, ((l + r) >> 1) + 1, r);
  }
  void build(vector<vector<long long>> &g, long long root) {
    long long sz = n;
    depth.resize(sz + 1);
    heavy.resize(sz + 1);
    par.resize(sz + 1);
    pos.resize(sz + 1);
    head.resize(sz + 1);
    sub.resize(sz + 1);
    seg.resize(4 * sz + 1);
    dfs(root, root, g);
    decompose(root, root, g);
    buildseg(1, 0, n - 1);
  }
  void dfs(long long i, long long p, vector<vector<long long>> &g) {
    depth[i] = depth[p] + 1;
    par[i] = p;
    sub[i] = 1;
    long long ms = 0, hc = -1;
    for (auto j : g[i]) {
      if (j == p) continue;
      dfs(j, i, g);
      sub[i] += sub[j];
      if (sub[j] > ms) {
        ms = sub[j];
        hc = j;
      }
    }
    heavy[i] = hc;
  }
  void decompose(long long i, long long h, vector<vector<long long>> &g) {
    pos[i] = dfsarray.size();
    dfsarray.push_back(i);
    head[i] = h;
    if (heavy[i] != -1) {
      decompose(heavy[i], h, g);
    }
    for (auto j : g[i]) {
      if (j == par[i] or j == heavy[i]) continue;
      decompose(j, j, g);
    }
  }
  long long lca(long long u, long long v) {
    for (; head[u] != head[v]; u = par[head[u]]) {
      if (depth[head[u]] < depth[head[v]]) {
        swap(u, v);
      }
    }
    return depth[u] > depth[v] ? v : u;
  }
  long long get(long long u, long long v) {
    long long ans = 0;
    long long sub = a[u] + a[v];
    for (; head[u] != head[v]; u = par[head[u]]) {
      if (depth[head[u]] < depth[head[v]]) {
        swap(u, v);
      }
      ans += query(pos[head[u]], pos[u]);
    }
    if (depth[u] > depth[v]) {
      swap(u, v);
    }
    ans += query(pos[u], pos[v]);
    ans *= 2;
    return ans - sub;
  }
};
vector<vector<long long>> g;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  g.resize(n + 1);
  for (long long i = 1; i < n; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  HEAVY_LIGHT h;
  h.build(g, 1);
  while (q--) {
    cin >> t >> x >> y;
    if (t == 1) {
      h.upd(h.pos[x], abs(y));
      a[x] = abs(y);
    } else {
      cout << h.get(x, y);
      cout << "\n";
    }
  }
}

```