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;
const int INF = 2147483647;
const int MAX = 100005;
struct BIT {
  int size;
  vector<int> arr;
  BIT(int s) {
    size = s;
    arr = vector<int>(s, 0);
  }
  void add(int pos, int val) {
    for (; pos < size; pos += pos & -pos) {
      arr[pos] += val;
    }
  }
  int sum(int pos) {
    int res = 0;
    for (; pos > 0; pos -= pos & -pos) {
      res += arr[pos];
    }
    return res;
  }
};
vector<vector<int> > v;
vector<int> adj[MAX];
pair<long long, long long> group[MAX];
vector<BIT> arr;
int main() {
  int n, q;
  cin >> n >> q;
  memset(adj, 0, sizeof(adj));
  int num = 0;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  group[0] = make_pair(0, 1);
  for (int i = 0; i < adj[0].size(); i++) {
    vector<int> v1;
    v1.push_back(0);
    int j = adj[0][i], last = 0;
    group[j] = make_pair(i + 1, 1);
    while (adj[j].size() == 2) {
      v1.push_back(adj[j][0] + adj[j][1] - j);
      int temp = j;
      j = adj[j][0] + adj[j][1] - last;
      last = temp;
      group[j] = make_pair(i + 1, v1.size());
    }
    v.push_back(v1);
  }
  BIT cen(MAX);
  arr.push_back(cen);
  for (int i = 1; i <= v.size(); i++) {
    BIT b(v[i - 1].size() + 5);
    arr.push_back(b);
  }
  for (int i = 0; i < q; i++) {
    int a;
    cin >> a;
    if (a == 0) {
      int b, c, d;
      cin >> b >> c >> d;
      b--;
      int x = group[b].first, y = group[b].second;
      if (x == 0) {
        arr[0].add(1, c);
        arr[0].add(min(MAX - 1, d + 2), -c);
      } else if (d >= y) {
        arr[x].add(min(arr[x].size - 1, d - y + 1), c);
        arr[x].add(min(arr[x].size - 1, y + d + 1), -c);
        arr[0].add(1, c);
        arr[0].add(min(MAX - 1, d - y + 2), -c);
      } else {
        arr[x].add(min(arr[x].size - 1, y - d), c);
        arr[x].add(min(arr[x].size - 1, y + d + 1), -c);
      }
    } else {
      int b;
      cin >> b;
      b--;
      int x = group[b].first, y = group[b].second;
      if (b == 0)
        cout << arr[0].sum(y) << endl;
      else
        cout << arr[0].sum(y + 1) + arr[x].sum(y) << endl;
    }
  }
}
