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 MAX = 2e5 + 9;
struct fen {
  map<int, int> f;
  void add(int p, int x) {
    for (p++; p < MAX; p += p & -p) f[p] += x;
  }
  int get(int p) {
    int ret = 0;
    for (; p; p -= p & -p) ret += f[p];
    return ret;
  }
} f[MAX], pub;
int col[MAX], c, a, b, n, q, h[MAX];
vector<int> g[MAX];
void dfs(int v, int p, int c) {
  col[v] = c;
  for (auto u : g[v])
    if (u != p) h[u] = h[v] + 1, dfs(u, v, c);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++)
    cin >> a >> b, a--, b--, g[a].push_back(b), g[b].push_back(a);
  for (auto i : g[0]) h[i] = 1, dfs(i, 0, ++c);
  while (q--) {
    int type, v, x, d;
    cin >> type >> v, v--;
    if (!type) {
      cin >> x >> d;
      if (h[v] <= d)
        f[col[v]].add(d - h[v] + 1, +x), f[col[v]].add(h[v] + d + 1, -x),
            pub.add(0, +x), pub.add(d - h[v] + 1, -x);
      else
        f[col[v]].add(h[v] - d, +x), f[col[v]].add(h[v] + d + 1, -x);
    } else
      cout << f[col[v]].get(h[v] + 1) + pub.get(h[v] + 1) << '\n';
  }
}
