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;
vector<int> G[100000 + 10], T[100000 + 10];
int n;
int Group[100000 + 10], dep[100000 + 10];
int lowbit(int x) { return x & (-x); }
void dfs(int depth, int root, int num, int pre) {
  Group[root] = num;
  dep[root] = depth;
  T[num].push_back(0);
  for (int i = 0; i < G[root].size(); i++) {
    if (pre == G[root][i]) continue;
    dfs(depth + 1, G[root][i], num, root);
  }
}
void modify(int num, int a, int b, int del) {
  int tt = T[num].size() - 1;
  b = min(b, tt);
  a = min(a - 1, tt);
  while (b) {
    T[num][b] += del;
    b = b - lowbit(b);
  }
  while (a) {
    T[num][a] -= del;
    a = a - lowbit(a);
  }
}
int query(int num, int x) {
  int sum = 0;
  while (x < T[num].size()) {
    sum = sum + T[num][x];
    x = x + lowbit(x);
  }
  return sum;
}
int main() {
  int q, i, a, b, x, c, root;
  while (scanf("%d%d", &n, &q) != EOF) {
    for (i = 1; i <= n; i++) G[i].clear();
    for (i = 1; i < n; i++) {
      scanf("%d%d", &a, &b);
      G[a].push_back(b);
      G[b].push_back(a);
    }
    int ttt = G[1].size();
    for (i = 1; i <= n; i++) {
      T[ttt].push_back(0);
    }
    for (i = 0; i < G[1].size(); i++) {
      T[i].push_back(0);
      dfs(1, G[1][i], i, 1);
    }
    root = 0;
    for (i = 1; i <= q; i++) {
      scanf("%d", &x);
      if (x) {
        scanf("%d", &a);
        if (a == 1)
          printf("%d\n", root);
        else {
          int ans = query(Group[a], dep[a]);
          int ans2 = query(G[1].size(), dep[a]);
          printf("%d\n", ans + ans2);
        }
      } else {
        scanf("%d%d%d", &a, &b, &c);
        if (a == 1) {
          root += b;
          modify(G[1].size(), 1, c, b);
        } else {
          if (c == dep[a]) {
            root += b;
            modify(Group[a], 1, c + dep[a], b);
          } else if (c < dep[a]) {
            modify(Group[a], dep[a] - c, dep[a] + c, b);
          } else if (c > dep[a]) {
            root += b;
            modify(Group[a], c - dep[a] + 1, c + dep[a], b);
            modify(G[1].size(), 1, c - dep[a], b);
          }
        }
      }
    }
  }
  return 0;
}
