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 N = 1e5 + 5;
vector<int> graph[N], t[N];
int inTree[N], level[N], n, q, x, dist, vt, a, b, cm, qt, inRoot;
void dfs(int u, int depth, int num, int parent) {
  inTree[u] = num;
  level[u] = depth;
  t[num].push_back(0);
  for (auto it : graph[u]) {
    if (it == parent) continue;
    dfs(it, depth + 1, num, u);
  }
}
void update(int pos, int val, int num) {
  int sz = t[num].size();
  for (int i = pos; i < sz; i += i & -i) t[num][i] += val;
}
int query(int pos, int num) {
  int res = 0;
  assert(pos < t[num].size());
  for (int i = pos; i > 0; i -= i & -i) {
    res += t[num][i];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    --a;
    --b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  cm = graph[0].size();
  for (int i = 0; i < cm; i++) {
    t[i].push_back(0);
    dfs(graph[0][i], 1, i, 0);
  }
  t[cm].resize(n + 10, 0);
  for (int i = 0; i < q; i++) {
    cin >> qt;
    if (qt == 0) {
      cin >> vt >> x >> dist;
      --vt;
      if (vt == 0) {
        inRoot += x;
        update(1, x, cm);
        update(dist + 1, -x, cm);
      } else {
        if (dist >= level[vt]) {
          int left = dist - level[vt];
          inRoot += x;
          update(1, x, cm);
          update(left + 1, -x, cm);
          update(left + 1, x, inTree[vt]);
          update(dist + level[vt] + 1, -x, inTree[vt]);
        } else {
          update(level[vt] - dist, x, inTree[vt]);
          update(level[vt] + dist + 1, -x, inTree[vt]);
        }
      }
    } else {
      cin >> vt;
      --vt;
      if (vt == 0) {
        cout << inRoot << "\n";
      } else {
        int ans = query(level[vt], inTree[vt]) + query(level[vt], cm);
        cout << ans << "\n";
      }
    }
  }
  return 0;
}
