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 aib[100010];
vector<int> adia[100010];
vector<vector<int>> aib_pers;
int ranc[100010], ramura[100010];
void dfs(int nod, int tata, int rnk, int rmr);
void adauga(int nod, int d, int x);
int sum(int nod);
int main() {
  ranc[1] = 1;
  int n, m, a, b;
  cin >> n >> m;
  n--;
  while (n--) {
    cin >> a >> b;
    adia[a].push_back(b);
    adia[b].push_back(a);
  }
  for (int i = 0; i < adia[1].size(); i++) {
    dfs(adia[1][i], 1, 2, i);
  }
  int q, x, v, d;
  while (m--) {
    cin >> q;
    if (q == 0) {
      cin >> v >> x >> d;
      adauga(v, d, x);
    } else {
      cin >> v;
      cout << sum(v) << '\n';
    }
  }
  return 0;
}
void dfs(int nod, int tata, int rnk, int rmr) {
  ranc[nod] = rnk;
  ramura[nod] = rmr;
  for (auto i : adia[nod]) {
    if (i != tata) {
      return dfs(i, nod, rnk + 1, rmr);
    }
  }
  aib_pers.push_back(vector<int>(rnk + 2, 0));
}
void adauga(int nod, int d, int x) {
  int n = d - ranc[nod] + 2, rmr = ramura[nod];
  if (n > 0) {
    int m = 1;
    while (m < 100010) {
      aib[m] += x;
      m += ((m) & (-(m)));
    }
    m = d - ranc[nod] + 3;
    while (m < 100010) {
      aib[m] -= x;
      m += ((m) & (-(m)));
    }
  }
  int l = ranc[nod] - d;
  int r = ranc[nod] + d + 1;
  if (l < n + 1) l = n + 1;
  if (l > r) return;
  while (l < aib_pers[rmr].size()) {
    aib_pers[rmr][l] += x;
    l += ((l) & (-(l)));
  }
  while (r < aib_pers[rmr].size()) {
    aib_pers[rmr][r] -= x;
    r += ((r) & (-(r)));
  }
}
int sum(int nod) {
  int k = ranc[nod], l = ramura[nod];
  int r = 0, n = k;
  while (n > 0) {
    r += aib[n];
    n -= ((n) & (-(n)));
  }
  if (nod != 1) {
    while (k > 0) {
      r += aib_pers[l][k];
      k -= ((k) & (-(k)));
    }
  }
  return r;
}
