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;
const int N = 100000;
vector<int> g[N];
vector<vector<int> > t;
int d[N], color[N], sz[N];
void dfs(int v, int p, int l, int c) {
  d[v] = l;
  color[v] = c;
  sz[c]++;
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    if (u != p) dfs(u, v, l + 1, v ? c : i + 1);
  }
}
void add(int v, int tl, int tr, int l, int r, int val, int c) {
  if (l > r) return;
  if (tl == l && tr == r)
    t[c][v] += val;
  else {
    int tm = (tl + tr) / 2;
    add(v * 2 + 1, tl, tm, l, min(tm, r), val, c);
    add(v * 2 + 2, tm + 1, tr, max(tm + 1, l), r, val, c);
  }
}
int get(int v, int tl, int tr, int pos, int c) {
  if (tl == tr) return t[c][v];
  int tm = (tl + tr) / 2;
  if (pos <= tm)
    return t[c][v] + get(v * 2 + 1, tl, tm, pos, c);
  else
    return t[c][v] + get(v * 2 + 2, tm + 1, tr, pos, c);
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    v--, u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0, -1, 0, 0);
  t.push_back(vector<int>(4 * N, 0));
  for (int i = 1; i < n && sz[i]; i++) t.push_back(vector<int>(4 * sz[i], 0));
  for (int i = 0; i < q; i++) {
    int cmd;
    cin >> cmd;
    if (cmd == 0) {
      int v, x, l;
      cin >> v >> x >> l;
      v--;
      int r = l - d[v];
      if (r >= 0) add(0, 0, N - 1, 0, r, x, 0);
      if (v && sz[color[v]] > r)
        add(0, 0, sz[color[v]] - 1, max(r, d[v] - 1 - l),
            min(d[v] - 1 + l, sz[color[v]] - 1), x, color[v]);
    } else {
      int v;
      cin >> v;
      v--;
      cout << get(0, 0, N - 1, d[v], 0) +
                  (v ? get(0, 0, sz[color[v]] - 1, d[v] - 1, color[v]) : 0)
           << endl;
    }
  }
  return 0;
}
