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 N = 1000 * 100 + 10;
vector<int> adj[N];
int h[N], en[N], st[N], pt, tr[N * 4][2], n;
bool mark[N];
void DFS(int v) {
  mark[v] = 1;
  st[v] = pt++;
  for (int u : adj[v]) {
    if (!mark[u]) {
      h[u] = h[v] + 1;
      DFS(u);
    }
  }
  en[v] = pt;
  return;
}
void add(int l, int r, bool f, int val, int b = 0, int e = n, int ind = 1) {
  if (l <= b && e <= r) {
    tr[ind][f] += val;
    return;
  }
  int mid = (b + e) / 2;
  if (l < mid) add(l, r, f, val, b, mid, ind * 2);
  if (r > mid) add(l, r, f, val, mid, e, ind * 2 + 1);
  return;
}
int get(int p, bool f, int b = 0, int e = n, int ind = 1) {
  int ret = tr[ind][f], mid = (b + e) / 2;
  if (b + 1 == e) return ret;
  if (p < mid)
    ret += get(p, f, b, mid, ind * 2);
  else
    ret += get(p, f, mid, e, ind * 2 + 1);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int q, u, v, tp, x, d;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  DFS(0);
  for (int i = 0; i < q; i++) {
    cin >> tp >> v;
    v--;
    if (tp)
      cout << get(st[v], 0) + get(h[v], 1) << endl;
    else {
      cin >> x >> d;
      int fs = st[v] - min(h[v] - 1, d);
      if (h[v] <= d) {
        int temp = d - h[v];
        add(0, temp + 1, 1, x);
        fs = max(fs, st[v] - (h[v] - temp) + 1);
      }
      int ls = min(en[v], st[v] + d + 1);
      if (fs < ls) add(fs, ls, 0, x);
    }
  }
  return 0;
}
