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[100005];
vector<int> it[100005];
int d[100005];
int tr[100005];
bool used[100005];
int n, q;
void dfs(int x, int dist, int tree) {
  d[x] = dist;
  used[x] = 1;
  tr[x] = tree;
  it[tree].push_back(0);
  for (int i = 0; i < g[x].size(); i++) {
    if (!used[g[x][i]]) {
      dfs(g[x][i], dist + 1, tree);
    }
  }
}
void add(int pos, int val, int tree) {
  if (pos >= it[tree].size()) pos = it[tree].size() - 1;
  for (int i = pos; i < it[tree].size(); i += i & (-i)) it[tree][i] += val;
}
int find(int pos, int tree) {
  int res = 0;
  for (int i = pos; i >= 1; i -= i & (-i)) res += it[tree][i];
  return res;
}
int main() {
  int x, y;
  int type, a, b, c;
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  d[1] = 0;
  tr[1] = 100001;
  it[tr[1]].resize(n + 10);
  used[1] = true;
  for (int i = 0; i < g[1].size(); i++) {
    it[i].push_back(0);
    dfs(g[1][i], 1, i);
    it[i].push_back(0);
  }
  int left;
  for (int i = 0; i < q; i++) {
    scanf("%d", &type);
    if (type == 0) {
      scanf("%d%d%d", &a, &b, &c);
      if (a == 1) {
        add(1, b, tr[a]);
        add(c + 2, -b, tr[a]);
      } else {
        if (d[a] - c > 0) {
          add(d[a] - c, b, tr[a]);
          add(d[a] + c + 1, -b, tr[a]);
        } else {
          left = c - d[a];
          add(1, b, tr[1]);
          add(left + 2, -b, tr[1]);
          add(left + 1, b, tr[a]);
          add(c + d[a] + 1, -b, tr[a]);
        }
      }
    } else {
      scanf("%d", &a);
      if (a == 1) {
        printf("%d\n", find(1, tr[a]));
      } else {
        printf("%d\n", find(d[a], tr[a]) + find(d[a] + 1, tr[1]));
      }
    }
  }
  return 0;
}
