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 = 200000 + 10;
const int inf = 0x3f3f3f3f;
vector<int> G[MAX];
vector<vector<int> > BIT;
int mark[MAX], dep[MAX];
int n, m, root;
void add_edge(int from, int to) {
  G[from].push_back(to);
  G[to].push_back(from);
}
void dfs(int k, int u, int pre, int d) {
  mark[u] = k;
  BIT[k].push_back(0);
  dep[u] = d;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v != pre) dfs(k, v, u, d + 1);
  }
  return;
}
int lowbit(int x) { return x & -x; }
void modify(int k, int x, int val) {
  for (int i = x; i < BIT[k].size(); i += lowbit(i)) BIT[k][i] += val;
}
int Query(int k, int x) {
  int res = 0;
  for (int i = x; i; i -= lowbit(i)) res += BIT[k][i];
  return res;
}
int main() {
  int a, b, t, v, x, d;
  scanf("%d %d", &n, &m);
  root = 0;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &a, &b);
    add_edge(a, b);
  }
  BIT.resize(G[1].size() + 100);
  for (int i = 0; i < G[1].size(); i++) {
    int v = G[1][i];
    dfs(i, v, 1, 1);
    BIT[i].push_back(0);
  }
  BIT[BIT.size() - 1].resize(n + 100);
  for (int i = 0; i < m; i++) {
    scanf("%d", &t);
    if (t == 0) {
      scanf("%d %d %d", &v, &x, &d);
      if (v == 1) {
        root += x;
        modify(BIT.size() - 1, 1, x);
        modify(BIT.size() - 1, d + 1, -x);
      } else {
        if (d >= dep[v]) {
          modify(BIT.size() - 1, 1, x);
          modify(BIT.size() - 1, (d - dep[v]) + 1, -x);
          root += x;
          modify(mark[v], (d - dep[v]) + 1, x);
          modify(mark[v], dep[v] + d + 1, -x);
        } else {
          modify(mark[v], dep[v] - d, x);
          modify(mark[v], dep[v] + d + 1, -x);
        }
      }
    } else {
      scanf("%d", &v);
      if (v == 1)
        printf("%d\n", root);
      else {
        printf("%d\n", Query(mark[v], dep[v]) + Query(BIT.size() - 1, dep[v]));
      }
    }
  }
  return 0;
}
