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;
long long fen[3][100010 + 10];
vector<int> pre;
int maxH[100010], pl[100010];
vector<int> edges[100010];
int dist[100010];
void add(int t, int x, int val) {
  if (t == 1)
    for (x++; x <= 100010; x += x & (-x)) fen[t][x] += val;
  else
    for (x++; x > 0; x -= x & (-x)) fen[t][x] += val;
}
long long get(int t, int x) {
  long long ans = 0;
  if (t == 1)
    for (x++; x > 0; x -= x & (-x)) ans += fen[t][x];
  else
    for (x++; x <= 100010; x += x & (-x)) ans += fen[t][x];
  return ans;
}
void DFS(int v, int f) {
  pre.push_back(v);
  pl[v] = pre.size() - 1;
  maxH[v] = 1;
  for (int i = 0; i < edges[v].size(); i++)
    if (edges[v][i] != f) {
      dist[edges[v][i]] = dist[v] + 1;
      DFS(edges[v][i], v);
      maxH[v] = maxH[edges[v][i]] + 1;
    }
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  DFS(1, -1);
  for (int i = 0; i < q; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int v;
      cin >> v;
      cout << get(0, dist[v]) + get(1, pl[v]) << endl;
    } else {
      int v, x, d;
      cin >> v >> x >> d;
      if (dist[v] <= d) {
        add(0, d - dist[v], x);
        add(1, pl[v] - dist[v] + 1, -x);
        add(1, min(pl[v] + maxH[v], pl[v] - dist[v] + 1 + d - dist[v]), x);
      }
      add(1, max(pl[v] - dist[v] + 1, pl[v] - d), x);
      add(1, min(pl[v] + maxH[v], pl[v] + d + 1), -x);
    }
  }
}
