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;
struct bit {
  int n;
  vector<long long> tree;
  bit(int n) : tree(n + 1, 0LL), n(n){};
  long long read(int idx) {
    long long sum = 0;
    while (idx > 0) {
      sum += tree[idx];
      idx -= (idx & -idx);
    }
    return sum;
  }
  long long read0(int idx) { return read(++idx); }
  void update(int idx, long long val) {
    while (idx <= n) {
      tree[idx] += val;
      idx += (idx & -idx);
    }
  }
  void update0(int idx, long long val) { update(++idx, val); }
  void update_range(int i, int j, long long val) {
    update(i, val);
    update(j + 1, -val);
  }
  void update_range0(int i, int j, long long val) {
    update0(i, val);
    update0(j + 1, -val);
  }
};
vector<vector<int> > g;
int level[100010], rama[100010], rsiz[100010], ramaCount;
void dfs(int v, int prev, int cnt) {
  level[v] = 1 + level[prev];
  rama[v] = cnt;
  if (g[v].size() == 1) rsiz[rama[v]] = level[v];
  for (int i = 0; i < g[v].size(); i++) {
    int next = g[v][i];
    if (next != prev) dfs(next, v, cnt);
  }
}
int main() {
  int n, q;
  cin >> n >> q;
  int x, y;
  g.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  ramaCount = 0;
  level[1] = 0;
  for (int i = 0; i < g[1].size(); i++) {
    int next = g[1][i];
    dfs(next, 1, ramaCount++);
  }
  vector<bit> rbits;
  for (int i = 0; i < ramaCount; i++) {
    rbits.push_back(bit(rsiz[i] + 10));
  }
  bit levelBit(5 + *max_element(level, level + n + 1));
  int t, v, d;
  long long X;
  for (int i = 0; i < q; i++) {
    cin >> t;
    if (!t) {
      cin >> v >> X >> d;
      if (v > 1) {
        rbits[rama[v]].update_range(max(1, level[v] - d),
                                    min(rsiz[rama[v]], level[v] + d), +X);
      }
      if (d >= level[v]) {
        levelBit.update_range0(0, d - level[v], +X);
        if (d > level[v] && v > 1) {
          rbits[rama[v]].update_range(1, min(rsiz[rama[v]], d - level[v]), -X);
        }
      }
    } else {
      cin >> v;
      long long ans = levelBit.read0(level[v]);
      if (v > 1) {
        ans += rbits[rama[v]].read(level[v]);
      }
      cout << ans << endl;
    }
  }
  return 0;
}
