#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
constexpr int DY[]{1, 0, -1, 0}, DX[]{0, -1, 0, 1};
constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1},
    DX8[]{0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;
struct HeavyLightDecomposition {
  std::vector<int> parent, subtree, id, inv, head;
  HeavyLightDecomposition(const std::vector<std::vector<int>> &graph,
                          int root = 0)
      : graph(graph) {
    int n = graph.size();
    parent.assign(n, -1);
    subtree.assign(n, 1);
    id.resize(n);
    inv.resize(n);
    head.resize(n);
    dfs1(root);
    head[root] = root;
    int now_id = 0;
    dfs2(root, now_id);
  }
  template <typename Fn>
  void v_update(int u, int v, Fn f) const {
    while (true) {
      if (id[u] > id[v]) std::swap(u, v);
      f(std::max(id[head[v]], id[u]), id[v] + 1);
      if (head[u] == head[v]) return;
      v = parent[head[v]];
    }
  }
  template <typename T, typename F, typename G>
  T v_query(int u, int v, F f, G g, const T ID) const {
    T left = ID, right = ID;
    while (true) {
      if (id[u] > id[v]) {
        std::swap(u, v);
        std::swap(left, right);
      }
      left = g(left, f(std::max(id[head[v]], id[u]), id[v] + 1));
      if (head[u] == head[v]) break;
      v = parent[head[v]];
    }
    return g(left, right);
  }
  template <typename Fn>
  void subtree_v_update(int ver, Fn f) const {
    f(id[ver], id[ver] + subtree[ver]);
  }
  template <typename T, typename Fn>
  T subtree_v_query(int ver, Fn f) const {
    return f(id[ver], id[ver] + subtree[ver]);
  }
  template <typename Fn>
  void e_update(int u, int v, Fn f) const {
    while (true) {
      if (id[u] > id[v]) std::swap(u, v);
      if (head[u] == head[v]) {
        f(id[u], id[v]);
        break;
      } else {
        f(id[head[v]] - 1, id[v]);
        v = parent[head[v]];
      }
    }
  }
  template <typename T, typename F, typename G>
  T e_query(int u, int v, F f, G g, const T ID) const {
    T left = ID, right = ID;
    while (true) {
      if (id[u] > id[v]) {
        std::swap(u, v);
        std::swap(left, right);
      }
      if (head[u] == head[v]) {
        left = g(left, f(id[u], id[v]));
        break;
      } else {
        left = g(left, f(id[head[v]] - 1, id[v]));
        v = parent[head[v]];
      }
    }
    return g(left, right);
  }
  template <typename Fn>
  void subtree_e_update(int ver, Fn f) const {
    f(id[ver], id[ver] + subtree[ver] - 1);
  }
  template <typename T, typename Fn>
  T subtree_e_query(int ver, Fn f) const {
    return f(id[ver], id[ver] + subtree[ver] - 1);
  }
  int lowest_common_ancestor(int u, int v) const {
    while (true) {
      if (id[u] > id[v]) std::swap(u, v);
      if (head[u] == head[v]) return u;
      v = parent[head[v]];
    }
  }

 private:
  std::vector<std::vector<int>> graph;
  void dfs1(int ver) {
    for (int &e : graph[ver]) {
      if (e != parent[ver]) {
        parent[e] = ver;
        dfs1(e);
        subtree[ver] += subtree[e];
        if (subtree[e] > subtree[graph[ver].front()])
          std::swap(e, graph[ver].front());
      }
    }
  }
  void dfs2(int ver, int &now_id) {
    id[ver] = now_id++;
    inv[id[ver]] = ver;
    for (int e : graph[ver]) {
      if (e != parent[ver]) {
        head[e] = (e == graph[ver].front() ? head[ver] : e);
        dfs2(e, now_id);
      }
    }
  }
};
template <typename Abelian>
struct FenwickTree {
  FenwickTree(int n, const Abelian ID = 0) : n(n), ID(ID), dat(n, ID) {}
  void add(int idx, Abelian val) {
    while (idx < n) {
      dat[idx] += val;
      idx |= idx + 1;
    }
  }
  Abelian sum(int idx) const {
    Abelian res = ID;
    --idx;
    while (idx >= 0) {
      res += dat[idx];
      idx = (idx & (idx + 1)) - 1;
    }
    return res;
  }
  Abelian sum(int left, int right) const {
    return left < right ? sum(right) - sum(left) : ID;
  }
  Abelian operator[](const int idx) const { return sum(idx, idx + 1); }
  int lower_bound(Abelian val) const {
    if (val <= ID) return 0;
    int res = 0, exponent = 1;
    while (exponent <= n) exponent <<= 1;
    for (int mask = exponent >> 1; mask > 0; mask >>= 1) {
      if (res + mask - 1 < n && dat[res + mask - 1] < val) {
        val -= dat[res + mask - 1];
        res += mask;
      }
    }
    return res;
  }

 private:
  int n;
  const Abelian ID;
  std::vector<Abelian> dat;
};
int main() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  vector<vector<int>> graph(n);
  for (int _ = (0); _ < (n - 1); ++_) {
    int s, t;
    cin >> s >> t;
    --s;
    --t;
    graph[s].emplace_back(t);
    graph[t].emplace_back(s);
  }
  HeavyLightDecomposition hld(graph);
  FenwickTree<ll> bit(n);
  for (int i = (0); i < (n); ++i) bit.add(hld.id[i], abs(a[i]));
  while (q--) {
    int type, u;
    cin >> type >> u;
    --u;
    if (type == 1) {
      int c;
      cin >> c;
      bit.add(hld.id[u], -abs(a[u]));
      a[u] = c;
      bit.add(hld.id[u], abs(a[u]));
    } else if (type == 2) {
      int v;
      cin >> v;
      --v;
      ll ans = hld.v_query(
          u, v, [&](int l, int r) -> ll { return bit.sum(l, r); },
          [](ll a, ll b) -> ll { return a + b; }, 0LL);
      ans *= 2;
      ans -= abs(a[u]);
      ans -= abs(a[v]);
      cout << ans << '\n';
    }
  }
}

### 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;
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
constexpr int DY[]{1, 0, -1, 0}, DX[]{0, -1, 0, 1};
constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1},
    DX8[]{0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;
struct HeavyLightDecomposition {
  std::vector<int> parent, subtree, id, inv, head;
  HeavyLightDecomposition(const std::vector<std::vector<int>> &graph,
                          int root = 0)
      : graph(graph) {
    int n = graph.size();
    parent.assign(n, -1);
    subtree.assign(n, 1);
    id.resize(n);
    inv.resize(n);
    head.resize(n);
    dfs1(root);
    head[root] = root;
    int now_id = 0;
    dfs2(root, now_id);
  }
  template <typename Fn>
  void v_update(int u, int v, Fn f) const {
    while (true) {
      if (id[u] > id[v]) std::swap(u, v);
      f(std::max(id[head[v]], id[u]), id[v] + 1);
      if (head[u] == head[v]) return;
      v = parent[head[v]];
    }
  }
  template <typename T, typename F, typename G>
  T v_query(int u, int v, F f, G g, const T ID) const {
    T left = ID, right = ID;
    while (true) {
      if (id[u] > id[v]) {
        std::swap(u, v);
        std::swap(left, right);
      }
      left = g(left, f(std::max(id[head[v]], id[u]), id[v] + 1));
      if (head[u] == head[v]) break;
      v = parent[head[v]];
    }
    return g(left, right);
  }
  template <typename Fn>
  void subtree_v_update(int ver, Fn f) const {
    f(id[ver], id[ver] + subtree[ver]);
  }
  template <typename T, typename Fn>
  T subtree_v_query(int ver, Fn f) const {
    return f(id[ver], id[ver] + subtree[ver]);
  }
  template <typename Fn>
  void e_update(int u, int v, Fn f) const {
    while (true) {
      if (id[u] > id[v]) std::swap(u, v);
      if (head[u] == head[v]) {
        f(id[u], id[v]);
        break;
      } else {
        f(id[head[v]] - 1, id[v]);
        v = parent[head[v]];
      }
    }
  }
  template <typename T, typename F, typename G>
  T e_query(int u, int v, F f, G g, const T ID) const {
    T left = ID, right = ID;
    while (true) {
      if (id[u] > id[v]) {
        std::swap(u, v);
        std::swap(left, right);
      }
      if (head[u] == head[v]) {
        left = g(left, f(id[u], id[v]));
        break;
      } else {
        left = g(left, f(id[head[v]] - 1, id[v]));
        v = parent[head[v]];
      }
    }
    return g(left, right);
  }
  template <typename Fn>
  void subtree_e_update(int ver, Fn f) const {
    f(id[ver], id[ver] + subtree[ver] - 1);
  }
  template <typename T, typename Fn>
  T subtree_e_query(int ver, Fn f) const {
    return f(id[ver], id[ver] + subtree[ver] - 1);
  }
  int lowest_common_ancestor(int u, int v) const {
    while (true) {
      if (id[u] > id[v]) std::swap(u, v);
      if (head[u] == head[v]) return u;
      v = parent[head[v]];
    }
  }

 private:
  std::vector<std::vector<int>> graph;
  void dfs1(int ver) {
    for (int &e : graph[ver]) {
      if (e != parent[ver]) {
        parent[e] = ver;
        dfs1(e);
        subtree[ver] += subtree[e];
        if (subtree[e] > subtree[graph[ver].front()])
          std::swap(e, graph[ver].front());
      }
    }
  }
  void dfs2(int ver, int &now_id) {
    id[ver] = now_id++;
    inv[id[ver]] = ver;
    for (int e : graph[ver]) {
      if (e != parent[ver]) {
        head[e] = (e == graph[ver].front() ? head[ver] : e);
        dfs2(e, now_id);
      }
    }
  }
};
template <typename Abelian>
struct FenwickTree {
  FenwickTree(int n, const Abelian ID = 0) : n(n), ID(ID), dat(n, ID) {}
  void add(int idx, Abelian val) {
    while (idx < n) {
      dat[idx] += val;
      idx |= idx + 1;
    }
  }
  Abelian sum(int idx) const {
    Abelian res = ID;
    --idx;
    while (idx >= 0) {
      res += dat[idx];
      idx = (idx & (idx + 1)) - 1;
    }
    return res;
  }
  Abelian sum(int left, int right) const {
    return left < right ? sum(right) - sum(left) : ID;
  }
  Abelian operator[](const int idx) const { return sum(idx, idx + 1); }
  int lower_bound(Abelian val) const {
    if (val <= ID) return 0;
    int res = 0, exponent = 1;
    while (exponent <= n) exponent <<= 1;
    for (int mask = exponent >> 1; mask > 0; mask >>= 1) {
      if (res + mask - 1 < n && dat[res + mask - 1] < val) {
        val -= dat[res + mask - 1];
        res += mask;
      }
    }
    return res;
  }

 private:
  int n;
  const Abelian ID;
  std::vector<Abelian> dat;
};
int main() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  vector<vector<int>> graph(n);
  for (int _ = (0); _ < (n - 1); ++_) {
    int s, t;
    cin >> s >> t;
    --s;
    --t;
    graph[s].emplace_back(t);
    graph[t].emplace_back(s);
  }
  HeavyLightDecomposition hld(graph);
  FenwickTree<ll> bit(n);
  for (int i = (0); i < (n); ++i) bit.add(hld.id[i], abs(a[i]));
  while (q--) {
    int type, u;
    cin >> type >> u;
    --u;
    if (type == 1) {
      int c;
      cin >> c;
      bit.add(hld.id[u], -abs(a[u]));
      a[u] = c;
      bit.add(hld.id[u], abs(a[u]));
    } else if (type == 2) {
      int v;
      cin >> v;
      --v;
      ll ans = hld.v_query(
          u, v, [&](int l, int r) -> ll { return bit.sum(l, r); },
          [](ll a, ll b) -> ll { return a + b; }, 0LL);
      ans *= 2;
      ans -= abs(a[u]);
      ans -= abs(a[v]);
      cout << ans << '\n';
    }
  }
}

```