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 maxn = 1e5 + 7;
vector<int> adj[maxn];
int fen[2][maxn];
int deg[maxn], bod[maxn], dod[maxn], lst[maxn];
int get(int idx, int* fen) {
  int sum = 0;
  while (idx > 0) {
    sum += fen[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void add(int idx, int val, int* fen) {
  if (!idx) return;
  while (idx < maxn) {
    fen[idx] += val;
    idx += (idx & -idx);
  }
}
int dfs(int v, int p = -1) {
  lst[v] = dod[v] + 1;
  for (int u : adj[v]) {
    if (u == p) continue;
    bod[u] = bod[v] + 1;
    dod[u] = lst[v];
    lst[v] = dfs(u, v);
  }
  return lst[v]--;
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  int src = 0;
  for (int i = 1; i < n; ++i)
    if (deg[i] > deg[src]) src = i;
  bod[src] = dod[src] = 1;
  dfs(src, -1);
  for (int i = 0; i < q; ++i) {
    int type;
    cin >> type;
    if (!type) {
      int v, x, d;
      cin >> v >> x >> d;
      v--;
      if (bod[v] - 1 > d) {
        add(dod[v] - d, x, fen[0]);
        add(min(dod[v] + d, lst[v]) + 1, -x, fen[0]);
      } else {
        add(d - bod[v] + 3, -x, fen[1]);
        add(1, x, fen[1]);
        if (v != src) {
          add(min(lst[v] + 1, dod[v] - 2 * bod[v] + d + 3), x, fen[0]);
          add(min(dod[v] + d, lst[v]) + 1, -x, fen[0]);
        }
      }
    } else {
      int v;
      cin >> v;
      v--;
      cout << get(dod[v], fen[0]) + get(bod[v], fen[1]) << "\n";
    }
  }
  return 0;
}
