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 maxn = 1e5 + 10, maxT = 1e5 + 10, Msk = (1 << 17), inf = 1e9 + 10,
          MOD = 1e9 + 7, Sq = 708;
const long long INF = 1e18 + 10;
vector<int> g[maxn];
int fen[maxn], h[maxn], p[maxn];
map<int, int> fn[maxn];
void dfs(int v, int par, int num, int dis) {
  h[v] = dis;
  p[v] = num;
  for (int u : g[v])
    if (u != par) dfs(u, v, num, dis + 1);
}
void add(int ind, int x) {
  for (; ind < maxn; ind += ind & -ind) fen[ind] += x;
}
void ad(int q, int ind, int x) {
  for (; ind < maxn; ind += ind & -ind) fn[q][ind] += x;
}
int query_1(int ind) {
  int ret = 0;
  for (; ind; ind -= ind & -ind) ret += fen[ind];
  return ret;
}
int query_2(int q, int ind) {
  int ret = 0;
  for (; ind; ind -= ind & -ind) ret += fn[q][ind];
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, q;
  h[1] = 1;
  cin >> n >> q;
  for (int i = 1, x, y; i < n; i++) {
    cin >> x >> y;
    g[x].push_back(y), g[y].push_back(x);
  }
  for (int i = 0; i < g[1].size(); i++) dfs(g[1][i], 1, i + 1, 2);
  for (int i = 0, type, v, x, d; i < q; i++) {
    cin >> type >> v;
    if (!type) {
      cin >> x >> d;
      if (v == 0) {
        add(1, x);
        add(d + 2, -x);
      } else {
        if (h[v] - d <= 1) {
          add(1, x);
          add(3 + d - h[v], -x);
          ad(p[v], 3 + d - h[v], x);
          ad(p[v], h[v] + d + 1, -x);
        } else {
          ad(p[v], h[v] - d, x);
          ad(p[v], h[v] + d + 1, -x);
        }
      }
    } else
      cout << query_1(h[v]) + query_2(p[v], h[v]) << endl;
  }
  return 0;
}
