#include <bits/stdc++.h>
using namespace std;
struct segtree {
  vector<long long> v;
  long long size;
  void init(long long n) {
    size = 1;
    while (size < n) size *= 2;
    v.resize(size * 2, 0);
  }
  void set(long long x, long long val) { set(x, val, 0, 0, size); }
  void set(long long x, long long val, long long cur, long long lx,
           long long rx) {
    if (rx - lx == 1) {
      v[cur] = val;
    } else {
      long long mid = (lx + rx) / 2;
      if (x < mid) {
        set(x, val, 2 * cur + 1, lx, mid);
      } else {
        set(x, val, 2 * cur + 2, mid, rx);
      }
      v[cur] = v[2 * cur + 1] + v[2 * cur + 2];
    }
  }
  long long get(long long l, long long r) { return get(l, r, 0, 0, size); }
  long long get(long long l, long long r, long long cur, long long lx,
                long long rx) {
    if (l >= rx or r <= lx) {
      return (long long)0;
    }
    if (l <= lx and r >= rx) {
      return v[cur];
    }
    long long m = (lx + rx) / 2;
    return (get(l, r, 2 * cur + 1, lx, m) + get(l, r, 2 * cur + 2, m, rx));
  }
};
struct HLD {
  long long cnt;
  vector<long long> depth, sz, parent, tp, idx;
  segtree st;
  HLD(vector<vector<long long> > &gr, long long root) {
    long long n = gr.size();
    depth.resize(n, -1);
    st.init(n + 1);
    sz.resize(n, 0);
    parent.resize(n, 0);
    tp.resize(n, 0);
    idx.resize(n, 0);
    dfs_sz(gr, root, root);
    cnt = 0;
    dfs_hld(gr, root, root, root);
  }
  long long dfs_sz(vector<vector<long long> > &gr, long long cur,
                   long long par) {
    depth[cur] = depth[par] + 1;
    sz[cur] = 1;
    parent[cur] = par;
    for (auto x : gr[cur]) {
      if (x == par) continue;
      sz[cur] += dfs_sz(gr, x, cur);
    }
    return sz[cur];
  }
  void dfs_hld(vector<vector<long long> > &gr, long long cur, long long par,
               long long top) {
    tp[cur] = top;
    idx[cur] = cnt++;
    long long mx_sz = -1, mx_child = -1;
    for (auto x : gr[cur]) {
      if (x == par) continue;
      if (sz[x] > mx_sz) {
        mx_sz = sz[x];
        mx_child = x;
      }
    }
    if (mx_child == -1) return;
    dfs_hld(gr, mx_child, cur, top);
    for (auto x : gr[cur]) {
      if (x == par or x == mx_child) continue;
      dfs_hld(gr, x, cur, x);
    }
  }
  long long path(long long x, long long y) {
    long long ans =
        -((st.get(idx[x], idx[x] + 1) + st.get(idx[y], idx[y] + 1)) / 2ll);
    while (tp[x] != tp[y]) {
      if (depth[tp[x]] < depth[tp[y]]) swap(x, y);
      ans += st.get(idx[tp[x]], idx[x] + 1);
      x = parent[tp[x]];
    }
    if (depth[x] > depth[y]) swap(x, y);
    ans += st.get(idx[x], idx[y] + 1);
    return ans;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, q;
  cin >> n >> q;
  vector<vector<long long> > gr(n);
  vector<long long> a(n);
  for (auto &x : a) cin >> x;
  for (long long i = 0; i + 1 < n; ++i) {
    long long x, y;
    cin >> x >> y;
    --x;
    --y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  HLD h(gr, 0);
  for (long long i = 0; i < n; ++i) {
    h.st.set(h.idx[i], 2ll * abs(a[i]));
  }
  while (q--) {
    long long ch, x, y;
    cin >> ch >> x >> y;
    if (ch == 1) {
      h.st.set(h.idx[x - 1], 2ll * abs(y));
    } else {
      --x;
      --y;
      cout << h.path(x, y) << "\n";
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 segtree {
  vector<long long> v;
  long long size;
  void init(long long n) {
    size = 1;
    while (size < n) size *= 2;
    v.resize(size * 2, 0);
  }
  void set(long long x, long long val) { set(x, val, 0, 0, size); }
  void set(long long x, long long val, long long cur, long long lx,
           long long rx) {
    if (rx - lx == 1) {
      v[cur] = val;
    } else {
      long long mid = (lx + rx) / 2;
      if (x < mid) {
        set(x, val, 2 * cur + 1, lx, mid);
      } else {
        set(x, val, 2 * cur + 2, mid, rx);
      }
      v[cur] = v[2 * cur + 1] + v[2 * cur + 2];
    }
  }
  long long get(long long l, long long r) { return get(l, r, 0, 0, size); }
  long long get(long long l, long long r, long long cur, long long lx,
                long long rx) {
    if (l >= rx or r <= lx) {
      return (long long)0;
    }
    if (l <= lx and r >= rx) {
      return v[cur];
    }
    long long m = (lx + rx) / 2;
    return (get(l, r, 2 * cur + 1, lx, m) + get(l, r, 2 * cur + 2, m, rx));
  }
};
struct HLD {
  long long cnt;
  vector<long long> depth, sz, parent, tp, idx;
  segtree st;
  HLD(vector<vector<long long> > &gr, long long root) {
    long long n = gr.size();
    depth.resize(n, -1);
    st.init(n + 1);
    sz.resize(n, 0);
    parent.resize(n, 0);
    tp.resize(n, 0);
    idx.resize(n, 0);
    dfs_sz(gr, root, root);
    cnt = 0;
    dfs_hld(gr, root, root, root);
  }
  long long dfs_sz(vector<vector<long long> > &gr, long long cur,
                   long long par) {
    depth[cur] = depth[par] + 1;
    sz[cur] = 1;
    parent[cur] = par;
    for (auto x : gr[cur]) {
      if (x == par) continue;
      sz[cur] += dfs_sz(gr, x, cur);
    }
    return sz[cur];
  }
  void dfs_hld(vector<vector<long long> > &gr, long long cur, long long par,
               long long top) {
    tp[cur] = top;
    idx[cur] = cnt++;
    long long mx_sz = -1, mx_child = -1;
    for (auto x : gr[cur]) {
      if (x == par) continue;
      if (sz[x] > mx_sz) {
        mx_sz = sz[x];
        mx_child = x;
      }
    }
    if (mx_child == -1) return;
    dfs_hld(gr, mx_child, cur, top);
    for (auto x : gr[cur]) {
      if (x == par or x == mx_child) continue;
      dfs_hld(gr, x, cur, x);
    }
  }
  long long path(long long x, long long y) {
    long long ans =
        -((st.get(idx[x], idx[x] + 1) + st.get(idx[y], idx[y] + 1)) / 2ll);
    while (tp[x] != tp[y]) {
      if (depth[tp[x]] < depth[tp[y]]) swap(x, y);
      ans += st.get(idx[tp[x]], idx[x] + 1);
      x = parent[tp[x]];
    }
    if (depth[x] > depth[y]) swap(x, y);
    ans += st.get(idx[x], idx[y] + 1);
    return ans;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, q;
  cin >> n >> q;
  vector<vector<long long> > gr(n);
  vector<long long> a(n);
  for (auto &x : a) cin >> x;
  for (long long i = 0; i + 1 < n; ++i) {
    long long x, y;
    cin >> x >> y;
    --x;
    --y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  HLD h(gr, 0);
  for (long long i = 0; i < n; ++i) {
    h.st.set(h.idx[i], 2ll * abs(a[i]));
  }
  while (q--) {
    long long ch, x, y;
    cin >> ch >> x >> y;
    if (ch == 1) {
      h.st.set(h.idx[x - 1], 2ll * abs(y));
    } else {
      --x;
      --y;
      cout << h.path(x, y) << "\n";
    }
  }
  return 0;
}

```