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;
vector<int> g[110000];
int n, q, x[110000], y[110000], c[110000];
int md[110000], a[110000], sum = 0, tim = 0;
void add(int s[], int pos, int k) {
  if (pos == 0) return;
  while (pos <= n) {
    s[pos] += k;
    pos += (pos & -pos);
  }
}
int ask(int s[], int pos) {
  int res = s[0];
  while (pos > 0) {
    res += s[pos];
    pos -= (pos & -pos);
  }
  return res;
}
void upd(int v, int k, int d) {
  if (md[v] > d) {
    add(c, x[v] - d, k);
    add(c, min(y[v], x[v] + d) + 1, -k);
  } else {
    add(a, d - md[v], k);
    sum += k;
    if (d == md[v]) a[0] += k;
    add(c, min(y[v], x[v] - md[v] + d - md[v]) + 1, k);
    add(c, min(y[v], x[v] + d) + 1, -k);
  }
}
int query(int v) {
  if (v == 1) return sum;
  return sum - ask(a, md[v] - 1) + ask(c, x[v]);
}
int down(int v, int d, int fa) {
  x[v] = ++tim;
  md[v] = d;
  if (g[v].size() == 1) return y[v] = x[v];
  if (g[v][0] == fa) return y[v] = down(g[v][1], d + 1, v);
  return y[v] = down(g[v][0], d + 1, v);
}
int main() {
  cin >> n >> q;
  int com, u, v, d;
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  x[1] = y[1] = ++tim;
  for (int i = 0; i < g[1].size(); ++i) down(g[1][i], 1, 1);
  while (q--) {
    cin >> com;
    if (com) {
      cin >> v;
      cout << query(v) << endl;
    } else {
      cin >> v >> u >> d;
      upd(v, u, d);
    }
  }
  return 0;
}
