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 n, q, dep[100005], entrada[100005], saida[100005], ft_local[100005],
    ft_global[100005];
int nodo = 1;
vector<int> adj[100005];
void solve(int u, int p) {
  entrada[u] = nodo++;
  for (int v : adj[u])
    if (v != p) {
      dep[v] = dep[u] + 1;
      solve(v, u);
    }
  saida[u] = nodo - 1;
}
void update(int* tree, int pos, int x) {
  if (!pos) return;
  for (int i = pos; i <= n; i += i & (-i)) tree[i] += x;
}
void update(int esq, int dir, int x, int* tree) {
  if (esq > dir) return;
  update(tree, esq, x);
  update(tree, dir + 1, -x);
}
long long get(int* tree, int pos) {
  long long sum = 0;
  for (int i = pos; i > 0; i -= i & (-i)) sum += tree[i];
  return sum;
}
int main() {
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dep[1] = 1;
  solve(1, -1);
  while (q--) {
    int op, v;
    cin >> op >> v;
    if (op == 0) {
      int x, d;
      cin >> x >> d;
      int k = d - dep[v] + 2;
      int finish = min(entrada[v] + d, saida[v]);
      if (k > 1) {
        update(1, k, x, ft_global);
        update(entrada[v] - dep[v] + (k + 1), finish, x, ft_local);
      } else if (k == 1) {
        int start = entrada[v] - d + 1;
        update(start, finish, x, ft_local);
        update(1, 1, x, ft_global);
      } else {
        int start = entrada[v] - d;
        update(start, finish, x, ft_local);
      }
    } else {
      cout << get(ft_local, entrada[v]) + get(ft_global, dep[v]) << endl;
    }
  }
  return 0;
}
