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;
struct BIT {
  vector<int> a;
  BIT(int sz = 0) { a.resize(sz); }
  void modify(int p, const int off) {
    p = min(p, (int)a.size() - 1);
    for (; p > 0; p -= ((p) & -(p))) a[p] += off;
  }
  int sum(int p) {
    int tot(0);
    for (; p < a.size(); p += ((p) & -(p))) tot += a[p];
    return tot;
  }
};
int ROOT;
vector<int> dep, idx;
vector<BIT> t;
vector<vector<int> > tr;
void dfs(int root, int d, int p, const int ii) {
  t[ii].a.push_back(0);
  dep[root] = d, idx[root] = ii;
  for (int i = 0; i < tr[root].size(); ++i) {
    if (tr[root][i] != p) dfs(tr[root][i], d + 1, root, ii);
  }
}
int main() {
  int n, q;
  int a, b;
  int k;
  int p, v, x, d;
  scanf("%d%d", &n, &q);
  tr.resize(n + 1);
  idx.resize(n + 1);
  dep.resize(n + 1);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &a, &b);
    tr[a].push_back(b);
    tr[b].push_back(a);
  }
  k = tr[1].size();
  t.resize(k + 1);
  for (int i = 0; i < tr[1].size(); ++i) {
    t[i].a.push_back(0);
    dfs(tr[1][i], 1, 1, i);
  }
  t[k].a.resize(n + 1);
  while (q--) {
    scanf("%d", &p);
    if (p) {
      scanf("%d", &v);
      if (v - 1)
        printf("%d\n", t[idx[v]].sum(dep[v]) + t[k].sum(dep[v]));
      else
        printf("%d\n", ROOT);
    } else {
      scanf("%d%d%d", &v, &x, &d);
      if (v - 1) {
        if (d >= dep[v]) {
          ROOT += x;
          t[k].modify(d - dep[v], x);
          t[idx[v]].modify(d - dep[v], -x);
        } else {
          t[idx[v]].modify(dep[v] - d - 1, -x);
        }
        t[idx[v]].modify(d + dep[v], x);
      } else {
        ROOT += x;
        t[k].modify(d, x);
      }
    }
  }
  return 0;
}
