#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4,avx2,abm,fma,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long INF = 0x3f3f3f3f;
const long long inf = pow(10, 18);
long long modulo = pow(10, 9) + 7;
long long valn[100100];
long long val[100100];
long long dp[100100];
vector<long long> sorted[100100];
vector<long long> g[100100];
long long lca[100100][20];
long long dep[100100];
long long tin[100100];
long long tout[100100];
long long calcval(long long a, long long b) {
  return max(abs(a + b), abs(a - b));
}
void dfslca(long long hd, long long pr, long long& tim) {
  lca[hd][0] = pr;
  tin[hd] = tim;
  for (auto hr : g[hd]) {
    if (hr == pr) continue;
    dep[hr] = dep[hd] + 1;
    ++tim;
    sorted[hd].push_back(hr);
    dfslca(hr, hd, tim);
  }
  tout[hd] = tim;
}
void pre(long long n) {
  for (long long j = 1; j < 20; j++)
    for (long long i = 0; i < n; i++) lca[i][j] = lca[lca[i][j - 1]][j - 1];
}
long long LCA(long long a, long long b) {
  if (dep[a] < dep[b]) swap(a, b);
  for (long long j = 20 - 1; j > -1; j--)
    if (dep[lca[a][j]] > dep[b]) a = lca[a][j];
  if (dep[a] != dep[b]) a = lca[a][0];
  for (long long j = 20 - 1; j > -1; j--)
    if (lca[a][j] != lca[b][j]) a = lca[a][j], b = lca[b][j];
  if (a != b) a = lca[a][0], b = lca[b][0];
  return a;
}
void calcdp(long long hd, long long pr) {
  for (auto& hr : g[hd]) {
    if (hr == pr) {
      continue;
    }
    dp[hr] = dp[hd] + calcval(val[hd], val[hr]);
    calcdp(hr, hd);
  }
}
bool isp(long long pr, long long hd) {
  return tin[pr] <= tin[hd] && tout[hd] <= tout[pr];
}
long long findChild(long long pr, long long hd) {
  long long lo = 0, hi = (long long)sorted[pr].size() - 1;
  while (lo < hi) {
    long long mid = (lo + hi) / 2 + 1;
    long long hr = sorted[pr][mid];
    if (tin[hr] > tin[hd]) {
      hi = mid - 1;
    } else {
      lo = mid;
    }
  }
  return sorted[pr][lo];
}
void deal() {
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < n; i++) {
    cin >> val[i];
    valn[i] = val[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    long long ai, bi;
    cin >> ai >> bi;
    --ai, --bi;
    g[ai].push_back(bi);
    g[bi].push_back(ai);
  }
  {
    long long tim = 1;
    dfslca(0, 0, tim);
    pre(n);
    calcdp(0, 0);
  }
  map<long long, long long> dif;
  for (long long l = 0; l < q; l++) {
    long long ty;
    cin >> ty;
    if (ty == 1) {
      long long u, c;
      cin >> u >> c;
      --u;
      dif[u] = c;
      valn[u] = c;
    } else {
      long long u, v;
      cin >> u >> v;
      --u, --v;
      long long lc = LCA(u, v);
      long long ans = dp[u] + dp[v] - 2 * dp[lc];
      unordered_set<long long> again;
      for (auto& el : dif) {
        long long pr = el.first;
        long long p1 = isp(pr, u);
        long long p2 = isp(pr, v);
        if (p1 && (!p2 || pr == lc) && pr != u) {
          long long a = pr;
          long long b = findChild(pr, u);
          if (b < a) {
            swap(a, b);
          }
          again.insert(a * 100100 + b);
        }
        if (p2 && (!p1 || pr == lc) && pr != v) {
          long long a = pr;
          long long b = findChild(pr, v);
          if (b < a) {
            swap(a, b);
          }
          again.insert(a * 100100 + b);
        }
        if (lca[pr][0] != pr &&
            ((p1 ^ p2) || (lca[pr][0] == lc && (p1 || p2)))) {
          long long a = pr;
          long long b = lca[pr][0];
          if (b < a) {
            swap(a, b);
          }
          again.insert(a * 100100 + b);
        }
      }
      for (auto& el : again) {
        long long a = el / 100100, b = el % 100100;
        ans -= calcval(val[a], val[b]);
        ans += calcval(valn[a], valn[b]);
      }
      cout << ans << '\n';
    }
    if ((long long)dif.size() > 300) {
      for (long long i = 0; i < n; i++) {
        val[i] = valn[i];
      }
      calcdp(0, 0);
      dif.clear();
    }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  deal();
}

### Prompt

Develop a solution in CPP 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4,avx2,abm,fma,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long INF = 0x3f3f3f3f;
const long long inf = pow(10, 18);
long long modulo = pow(10, 9) + 7;
long long valn[100100];
long long val[100100];
long long dp[100100];
vector<long long> sorted[100100];
vector<long long> g[100100];
long long lca[100100][20];
long long dep[100100];
long long tin[100100];
long long tout[100100];
long long calcval(long long a, long long b) {
  return max(abs(a + b), abs(a - b));
}
void dfslca(long long hd, long long pr, long long& tim) {
  lca[hd][0] = pr;
  tin[hd] = tim;
  for (auto hr : g[hd]) {
    if (hr == pr) continue;
    dep[hr] = dep[hd] + 1;
    ++tim;
    sorted[hd].push_back(hr);
    dfslca(hr, hd, tim);
  }
  tout[hd] = tim;
}
void pre(long long n) {
  for (long long j = 1; j < 20; j++)
    for (long long i = 0; i < n; i++) lca[i][j] = lca[lca[i][j - 1]][j - 1];
}
long long LCA(long long a, long long b) {
  if (dep[a] < dep[b]) swap(a, b);
  for (long long j = 20 - 1; j > -1; j--)
    if (dep[lca[a][j]] > dep[b]) a = lca[a][j];
  if (dep[a] != dep[b]) a = lca[a][0];
  for (long long j = 20 - 1; j > -1; j--)
    if (lca[a][j] != lca[b][j]) a = lca[a][j], b = lca[b][j];
  if (a != b) a = lca[a][0], b = lca[b][0];
  return a;
}
void calcdp(long long hd, long long pr) {
  for (auto& hr : g[hd]) {
    if (hr == pr) {
      continue;
    }
    dp[hr] = dp[hd] + calcval(val[hd], val[hr]);
    calcdp(hr, hd);
  }
}
bool isp(long long pr, long long hd) {
  return tin[pr] <= tin[hd] && tout[hd] <= tout[pr];
}
long long findChild(long long pr, long long hd) {
  long long lo = 0, hi = (long long)sorted[pr].size() - 1;
  while (lo < hi) {
    long long mid = (lo + hi) / 2 + 1;
    long long hr = sorted[pr][mid];
    if (tin[hr] > tin[hd]) {
      hi = mid - 1;
    } else {
      lo = mid;
    }
  }
  return sorted[pr][lo];
}
void deal() {
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < n; i++) {
    cin >> val[i];
    valn[i] = val[i];
  }
  for (long long i = 0; i < n - 1; i++) {
    long long ai, bi;
    cin >> ai >> bi;
    --ai, --bi;
    g[ai].push_back(bi);
    g[bi].push_back(ai);
  }
  {
    long long tim = 1;
    dfslca(0, 0, tim);
    pre(n);
    calcdp(0, 0);
  }
  map<long long, long long> dif;
  for (long long l = 0; l < q; l++) {
    long long ty;
    cin >> ty;
    if (ty == 1) {
      long long u, c;
      cin >> u >> c;
      --u;
      dif[u] = c;
      valn[u] = c;
    } else {
      long long u, v;
      cin >> u >> v;
      --u, --v;
      long long lc = LCA(u, v);
      long long ans = dp[u] + dp[v] - 2 * dp[lc];
      unordered_set<long long> again;
      for (auto& el : dif) {
        long long pr = el.first;
        long long p1 = isp(pr, u);
        long long p2 = isp(pr, v);
        if (p1 && (!p2 || pr == lc) && pr != u) {
          long long a = pr;
          long long b = findChild(pr, u);
          if (b < a) {
            swap(a, b);
          }
          again.insert(a * 100100 + b);
        }
        if (p2 && (!p1 || pr == lc) && pr != v) {
          long long a = pr;
          long long b = findChild(pr, v);
          if (b < a) {
            swap(a, b);
          }
          again.insert(a * 100100 + b);
        }
        if (lca[pr][0] != pr &&
            ((p1 ^ p2) || (lca[pr][0] == lc && (p1 || p2)))) {
          long long a = pr;
          long long b = lca[pr][0];
          if (b < a) {
            swap(a, b);
          }
          again.insert(a * 100100 + b);
        }
      }
      for (auto& el : again) {
        long long a = el / 100100, b = el % 100100;
        ans -= calcval(val[a], val[b]);
        ans += calcval(valn[a], valn[b]);
      }
      cout << ans << '\n';
    }
    if ((long long)dif.size() > 300) {
      for (long long i = 0; i < n; i++) {
        val[i] = valn[i];
      }
      calcdp(0, 0);
      dif.clear();
    }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  deal();
}

```