#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() { return abs((int)mt()); }
inline int mrand(int k) { return abs((int)mt()) % k; }
namespace HLD {
int n;
vector<vector<int>> adj;
vector<int> size;
vector<int> lev;
vector<int> p;
vector<int> heavy;
vector<int> num;
vector<int> head;
int cnt;
void firstdfs(int u, int dad) {
  size[u] = 1, heavy[u] = -1;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v != dad) {
      p[v] = u, lev[v] = lev[u] + 1;
      firstdfs(v, u);
      size[u] += size[v];
      if (heavy[u] == -1 || size[heavy[u]] < size[v]) {
        heavy[u] = v;
      }
    }
  }
}
void dfs(int u, int h, int p) {
  num[u] = cnt++, head[u] = h;
  if (~heavy[u]) {
    dfs(heavy[u], h, u);
  }
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v != p && v != heavy[u]) {
      dfs(v, v, u);
    }
  }
}
void build(const vector<vector<int>>& _adj) {
  n = (int)_adj.size();
  adj = _adj;
  size.resize(n);
  lev.resize(n);
  p.resize(n);
  heavy.resize(n);
  num.resize(n);
  head.resize(n);
  cnt = 0;
  firstdfs(0, -1), dfs(0, 0, -1);
}
vector<pair<int, int>> getsegs(int u, int v) {
  vector<pair<int, int>> res;
  int hu = head[u], hv = head[v];
  while (hu != hv) {
    if (lev[hu] < lev[hv]) swap(u, v), swap(hu, hv);
    res.push_back(make_pair(num[hu], num[u]));
    u = p[hu], hu = head[u];
  }
  if (lev[u] > lev[v]) swap(u, v);
  res.push_back(make_pair(num[u], num[v]));
  return res;
}
}  // namespace HLD
void chemthan() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i], a[i] = abs(a[i]);
  vector<vector<int>> adj(n);
  map<pair<int, int>, int> hs;
  for (int i = (0); i < (n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v), adj[v].push_back(u);
    hs[make_pair(u, v)] = hs[make_pair(v, u)] = 1;
  }
  HLD::build(adj);
  vector<long long> fen(n + 1);
  auto upd = [&](int p, int v) {
    p++;
    for (; p < int((fen).size()); p += p & -p) {
      fen[p] += v;
    }
  };
  auto query = [&](int p) {
    p++;
    long long res = 0;
    for (; 0 < p; p -= p & -p) {
      res += fen[p];
    }
    return res;
  };
  for (int i = (0); i < (n); ++i) {
    upd(HLD::num[i], a[i]);
  }
  while (q--) {
    int op, u, v;
    cin >> op >> u >> v;
    if (op == 1) {
      u--, v = abs(v);
      upd(HLD::num[u], -a[u]);
      a[u] = v;
      upd(HLD::num[u], a[u]);
    } else {
      u--, v--;
      if (u == v) {
        cout << 0 << "\n";
        continue;
      }
      long long res = 0;
      for (auto [l, r] : HLD::getsegs(u, v)) {
        res += query(r);
        res -= query(l - 1);
      }
      if (!hs[make_pair(u, v)]) {
        res *= 2;
        res -= a[u];
        res -= a[v];
      }
      cout << res << "\n";
    }
  }
}
int32_t main(int32_t argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() { return abs((int)mt()); }
inline int mrand(int k) { return abs((int)mt()) % k; }
namespace HLD {
int n;
vector<vector<int>> adj;
vector<int> size;
vector<int> lev;
vector<int> p;
vector<int> heavy;
vector<int> num;
vector<int> head;
int cnt;
void firstdfs(int u, int dad) {
  size[u] = 1, heavy[u] = -1;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v != dad) {
      p[v] = u, lev[v] = lev[u] + 1;
      firstdfs(v, u);
      size[u] += size[v];
      if (heavy[u] == -1 || size[heavy[u]] < size[v]) {
        heavy[u] = v;
      }
    }
  }
}
void dfs(int u, int h, int p) {
  num[u] = cnt++, head[u] = h;
  if (~heavy[u]) {
    dfs(heavy[u], h, u);
  }
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v != p && v != heavy[u]) {
      dfs(v, v, u);
    }
  }
}
void build(const vector<vector<int>>& _adj) {
  n = (int)_adj.size();
  adj = _adj;
  size.resize(n);
  lev.resize(n);
  p.resize(n);
  heavy.resize(n);
  num.resize(n);
  head.resize(n);
  cnt = 0;
  firstdfs(0, -1), dfs(0, 0, -1);
}
vector<pair<int, int>> getsegs(int u, int v) {
  vector<pair<int, int>> res;
  int hu = head[u], hv = head[v];
  while (hu != hv) {
    if (lev[hu] < lev[hv]) swap(u, v), swap(hu, hv);
    res.push_back(make_pair(num[hu], num[u]));
    u = p[hu], hu = head[u];
  }
  if (lev[u] > lev[v]) swap(u, v);
  res.push_back(make_pair(num[u], num[v]));
  return res;
}
}  // namespace HLD
void chemthan() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i], a[i] = abs(a[i]);
  vector<vector<int>> adj(n);
  map<pair<int, int>, int> hs;
  for (int i = (0); i < (n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v), adj[v].push_back(u);
    hs[make_pair(u, v)] = hs[make_pair(v, u)] = 1;
  }
  HLD::build(adj);
  vector<long long> fen(n + 1);
  auto upd = [&](int p, int v) {
    p++;
    for (; p < int((fen).size()); p += p & -p) {
      fen[p] += v;
    }
  };
  auto query = [&](int p) {
    p++;
    long long res = 0;
    for (; 0 < p; p -= p & -p) {
      res += fen[p];
    }
    return res;
  };
  for (int i = (0); i < (n); ++i) {
    upd(HLD::num[i], a[i]);
  }
  while (q--) {
    int op, u, v;
    cin >> op >> u >> v;
    if (op == 1) {
      u--, v = abs(v);
      upd(HLD::num[u], -a[u]);
      a[u] = v;
      upd(HLD::num[u], a[u]);
    } else {
      u--, v--;
      if (u == v) {
        cout << 0 << "\n";
        continue;
      }
      long long res = 0;
      for (auto [l, r] : HLD::getsegs(u, v)) {
        res += query(r);
        res -= query(l - 1);
      }
      if (!hs[make_pair(u, v)]) {
        res *= 2;
        res -= a[u];
        res -= a[v];
      }
      cout << res << "\n";
    }
  }
}
int32_t main(int32_t argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

```