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<vector<int> > bit;
vector<int> bitn;
vector<int> e[100000];
int num[100000];
int dep[100000];
void add(int i, int j, int x) {
  while (j <= bitn[i]) {
    bit[i][j] += x;
    j += j & -j;
  }
}
int sum(int i, int j) {
  int s = 0;
  while (j > 0) {
    s += bit[i][j];
    j -= j & -j;
  }
  return s;
}
void dfs(int v, int p, int k, int d) {
  num[v] = k;
  dep[v] = d;
  bit[k].push_back(0);
  for (int i = 0; i < e[v].size(); i++) {
    if (e[v][i] == p) continue;
    dfs(e[v][i], v, k, d + 1);
  }
}
int main() {
  int n, q;
  int i, j;
  scanf("%d %d", &n, &q);
  for (i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dep[0] = 0;
  bit.push_back(vector<int>());
  for (i = 0; i <= n + 5; i++) bit[0].push_back(0);
  bitn.push_back(n + 5);
  for (i = 0; i < e[0].size(); i++) {
    bit.push_back(vector<int>());
    dfs(e[0][i], 0, i + 1, 1);
    bitn.push_back(bit[i + 1].size());
    bit[i + 1].push_back(0);
  }
  for (i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 0) {
      int v, x, d;
      scanf("%d %d %d", &v, &x, &d);
      v--;
      if (v == 0) {
        add(0, 1, x);
        add(0, d + 2, -x);
      } else {
        int k = num[v];
        if (dep[v] <= d) {
          add(k, d - dep[v] + 1, x);
          add(k, dep[v] + d + 1, -x);
          add(0, 1, x);
          add(0, d - dep[v] + 2, -x);
        } else {
          add(k, dep[v] - d, x);
          add(k, dep[v] + d + 1, -x);
        }
      }
    } else {
      int v;
      scanf("%d", &v);
      v--;
      if (v == 0)
        printf("%d\n", sum(0, 1));
      else {
        int d = dep[v];
        int k = num[v];
        int s = sum(k, d) + sum(0, d + 1);
        printf("%d\n", s);
      }
    }
  }
  return 0;
}
