#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 100005;
int n, qq;
int a[N];
vector<int> g[N];
int q[N];
int p0[N];
void dfs0(int x, int p) {
  p0[x] = p;
  q[x] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) continue;
    dfs0(h, x);
    q[x] += q[h];
  }
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) swap(g[x][i], g[x].back());
  }
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) continue;
    if (q[h] > q[g[x][0]]) swap(g[x][i], g[x][0]);
  }
}
int f[N];
int tin[N], tout[N], ti;
void dfs1(int x, int p) {
  tin[x] = ++ti;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) continue;
    if (i == 0) {
      f[h] = f[x];
      dfs1(h, x);
    } else {
      f[h] = h;
      dfs1(h, x);
    }
  }
  tout[x] = ti;
}
long long t[N * 4];
void ubd(int tl, int tr, int x, int y, int pos) {
  if (tl == tr) {
    t[pos] = y;
    return;
  }
  int m = (tl + tr) / 2;
  if (x <= m)
    ubd(tl, m, x, y, pos * 2);
  else
    ubd(m + 1, tr, x, y, pos * 2 + 1);
  t[pos] = t[pos * 2] + t[pos * 2 + 1];
}
long long qry(int tl, int tr, int l, int r, int pos) {
  if (l > r) return 0;
  if (tl == l && tr == r) return t[pos];
  int m = (tl + tr) / 2;
  return (qry(tl, m, l, min(m, r), pos * 2) +
          qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
}
long long qryy(int x, int y) {
  long long ans = 0;
  while (1) {
    if (tin[x] < tin[y]) swap(x, y);
    if (f[x] == f[y]) {
      ans += qry(1, n, tin[y], tin[x], 1);
      return ans;
    }
    ans += qry(1, n, tin[f[x]], tin[x], 1);
    x = p0[f[x]];
  }
}
void solv() {
  cin >> n >> qq;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs0(1, 1);
  f[1] = 1;
  dfs1(1, 1);
  for (int x = 1; x <= n; ++x) ubd(1, n, tin[x], a[x], 1);
  while (qq--) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int x, y;
      cin >> x >> y;
      a[x] = y;
      a[x] = abs(a[x]);
      ubd(1, n, tin[x], a[x], 1);
    } else {
      int x, y;
      cin >> x >> y;
      cout << qryy(x, y) * 2 - a[x] - a[y] << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  while (tt--) solv();
  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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 100005;
int n, qq;
int a[N];
vector<int> g[N];
int q[N];
int p0[N];
void dfs0(int x, int p) {
  p0[x] = p;
  q[x] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) continue;
    dfs0(h, x);
    q[x] += q[h];
  }
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) swap(g[x][i], g[x].back());
  }
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) continue;
    if (q[h] > q[g[x][0]]) swap(g[x][i], g[x][0]);
  }
}
int f[N];
int tin[N], tout[N], ti;
void dfs1(int x, int p) {
  tin[x] = ++ti;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) continue;
    if (i == 0) {
      f[h] = f[x];
      dfs1(h, x);
    } else {
      f[h] = h;
      dfs1(h, x);
    }
  }
  tout[x] = ti;
}
long long t[N * 4];
void ubd(int tl, int tr, int x, int y, int pos) {
  if (tl == tr) {
    t[pos] = y;
    return;
  }
  int m = (tl + tr) / 2;
  if (x <= m)
    ubd(tl, m, x, y, pos * 2);
  else
    ubd(m + 1, tr, x, y, pos * 2 + 1);
  t[pos] = t[pos * 2] + t[pos * 2 + 1];
}
long long qry(int tl, int tr, int l, int r, int pos) {
  if (l > r) return 0;
  if (tl == l && tr == r) return t[pos];
  int m = (tl + tr) / 2;
  return (qry(tl, m, l, min(m, r), pos * 2) +
          qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
}
long long qryy(int x, int y) {
  long long ans = 0;
  while (1) {
    if (tin[x] < tin[y]) swap(x, y);
    if (f[x] == f[y]) {
      ans += qry(1, n, tin[y], tin[x], 1);
      return ans;
    }
    ans += qry(1, n, tin[f[x]], tin[x], 1);
    x = p0[f[x]];
  }
}
void solv() {
  cin >> n >> qq;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] = abs(a[i]);
  }
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs0(1, 1);
  f[1] = 1;
  dfs1(1, 1);
  for (int x = 1; x <= n; ++x) ubd(1, n, tin[x], a[x], 1);
  while (qq--) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int x, y;
      cin >> x >> y;
      a[x] = y;
      a[x] = abs(a[x]);
      ubd(1, n, tin[x], a[x], 1);
    } else {
      int x, y;
      cin >> x >> y;
      cout << qryy(x, y) * 2 - a[x] - a[y] << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  while (tt--) solv();
  return 0;
}

```