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 n, dist[100100], branch[100100], sz[100100];
vector<int> a[100100];
int *f[100100], fromRoot[100100];
void visit(int x, int par, int idBranch) {
  branch[x] = idBranch;
  sz[idBranch]++;
  for (int i = 0; i < int(a[x].size()); i++) {
    int y = a[x][i];
    if (y == par) continue;
    dist[y] = dist[x] + 1;
    visit(y, x, idBranch);
  }
}
void add(int f[], int x, int val, int mx) {
  for (int i = x + 1; i <= mx + 1; i += i & -i) f[i] += val;
}
int get(int f[], int x) {
  int res = 0;
  for (int i = x + 1; i; i -= i & -i) res += f[i];
  return res;
}
int main() {
  int Q, x, y, type, v, d;
  cin >> n >> Q;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (int i = 0; i < int(a[1].size()); i++) {
    dist[a[1][i]] = 1;
    visit(a[1][i], 1, i);
    f[i] = new int[sz[i] + 5];
    for (int j = 0; j <= sz[i] + 1; j++) f[i][j] = 0;
  }
  while (Q--) {
    scanf("%d%d", &type, &v);
    if (type) {
      int ans = get(fromRoot, dist[v]);
      if (v != 1) ans += get(f[branch[v]], dist[v]);
      printf("%d\n", ans);
    } else {
      scanf("%d%d", &x, &d);
      if (v != 1) {
        int idBranch = branch[v];
        int from = max(1, dist[v] - d), to = min(sz[idBranch], dist[v] + d);
        d -= dist[v];
        from = max(from, d + 1);
        if (from <= to) {
          add(f[idBranch], from, x, sz[idBranch]);
          add(f[idBranch], to + 1, -x, sz[idBranch]);
        }
      }
      if (d < 0) continue;
      add(fromRoot, 0, x, n);
      add(fromRoot, d + 1, -x, n);
    }
  }
}
