A little girl loves problems on trees very much. Here's one of them.

A tree is an undirected connected graph, not containing cycles. The degree of node x in the tree is the number of nodes y of the tree, such that each of them is connected with node x by some edge of the tree. 

Let's consider a tree that consists of n nodes. We'll consider the tree's nodes indexed from 1 to n. The cosidered tree has the following property: each node except for node number 1 has the degree of at most 2.

Initially, each node of the tree contains number 0. Your task is to quickly process the requests of two types:

  * Request of form: 0 v x d. In reply to the request you should add x to all numbers that are written in the nodes that are located at the distance of at most d from node v. The distance between two nodes is the number of edges on the shortest path between them. 
  * Request of form: 1 v. In reply to the request you should print the current number that is written in node v. 

Input

The first line contains integers n (2 ≤ n ≤ 105) and q (1 ≤ q ≤ 105) — the number of tree nodes and the number of requests, correspondingly.

Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), that show that there is an edge between nodes ui and vi. Each edge's description occurs in the input exactly once. It is guaranteed that the given graph is a tree that has the property that is described in the statement.

Next q lines describe the requests.

  * The request to add has the following format: 0 v x d (1 ≤ v ≤ n, 1 ≤ x ≤ 104, 1 ≤ d < n). 
  * The request to print the node value has the following format: 1 v (1 ≤ v ≤ n). 



The numbers in the lines are separated by single spaces.

Output

For each request to print the node value print an integer — the reply to the request.

Examples

Input

3 6
1 2
1 3
0 3 1 2
0 2 3 1
0 1 5 2
1 1
1 2
1 3


Output

9
9
6


Input

6 11
1 2
2 5
5 4
1 6
1 3
0 3 1 3
0 3 4 5
0 2 1 4
0 1 5 5
0 4 6 2
1 1
1 2
1 3
1 4
1 5
1 6


Output

11
17
11
16
17
11
#include <bits/stdc++.h>
using namespace std;
int log2(int n) {
  int lg = 0, m = n;
  while (m > 1) m /= 2, lg++;
  return n > (1 << lg) ? lg + 1 : lg;
}
template <class T>
struct itree {
  itree(int n) : k(log2(n)), n(1 << this->k), a(new T[this->n * 2]()) {}
  itree(itree &&t) : k(t.k), n(t.n), a(move(t.a)) {}
  void modify(const T &x, int l, int r, int L, int R, int i) {
    if (l == L && r == R) {
      a[i] += x;
      assert(i < n * 2);
    } else {
      int M = (L + R) / 2;
      if (r <= M)
        modify(x, l, r, L, M, i * 2);
      else if (l >= M)
        modify(x, l, r, M, R, i * 2 + 1);
      else
        modify(x, l, M, L, M, i * 2), modify(x, M, r, M, R, i * 2 + 1);
    }
  }
  void modify(const T &x, int l, int r) {
    l = max<int>(0, l);
    r = min<int>(r, n);
    if (l < r) modify(x, l, r, 0, n, 1);
  }
  T get(int j, int L, int R, int i) {
    if (j == L && j + 1 == R)
      return a[i];
    else {
      int M = (L + R) / 2;
      if (j < M)
        return a[i] + get(j, L, M, i * 2);
      else
        return a[i] + get(j, M, R, i * 2 + 1);
    }
  }
  T get(int i) { return get(i, 0, n, 1); }
  int k, n;
  unique_ptr<T[]> a;
};
int main() {
  int n, q;
  cin >> n >> q;
  bool hack = false;
  struct V {
    V() : u(0), v(0), d(0), x(0), b(0) {}
    int u, v, d, x, b;
  };
  vector<V> g(n + 1);
  vector<int> b;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    if (g[u].u)
      g[u].v = v;
    else
      g[u].u = v;
    if (g[v].u)
      g[v].v = u;
    else
      g[v].u = u;
    if (u == 1)
      b.push_back(v);
    else if (v == 1)
      b.push_back(u);
  }
  itree<int> root(n);
  vector<itree<int>> branches;
  for (int i = 0; i < b.size(); i++) {
    int v = b[i];
    int bi = branches.size();
    g[v].b = bi;
    int d = 1;
    for (int next = v, cur = next, prev = 1; cur; prev = cur, cur = next, d++) {
      next = prev == g[cur].u ? g[cur].v : g[cur].u;
      g[cur].d = d;
      g[cur].b = bi;
    }
    branches.push_back(itree<int>(d));
  }
  for (int i = 0; i < q; i++) {
    int t, v, x, d;
    cin >> t;
    if (t) {
      cin >> v;
      int x = root.get(g[v].d);
      if (v != 1) x += branches[g[v].b].get(g[v].d);
      if (!hack) cout << x << endl;
    } else {
      cin >> v >> x >> d;
      int md = d - g[v].d;
      if (v != 1)
        branches[g[v].b].modify(x, max<int>(g[v].d - d, md + 1),
                                g[v].d + d + 1);
      if (md >= 0) root.modify(x, 0, md + 1);
    }
    if (hack && !(i % 0x400)) cout << i << endl;
  }
}
