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 MAX_N = 100000 + 20;
struct BIT {
  int size;
  vector<long long> fen;
  BIT() { size = 0; }
  BIT(int n) {
    size = n;
    fen.resize(n + 1);
  }
  void add(int x, int val) {
    for (; x <= size; x += (x) & (-x)) fen[x] += val;
  }
  long long get(int x) {
    long long val = 0;
    for (; x > 0; x -= (x) & (-x)) val += fen[x];
    return val;
  }
};
vector<int> adj[MAX_N];
vector<BIT> a;
int mark[MAX_N], lvl[MAX_N];
int n, Q, head = 1;
long long rootAns;
int dfs(int v, int col) {
  mark[v] = col;
  int res = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (mark[u]) continue;
    lvl[u] = lvl[v] + 1;
    res += dfs(u, col);
  }
  return res;
}
void addInterVal(int x, int l, int r, int val) {
  if (l > r || !l || !r) return;
  a[x].add(l, val);
  a[x].add(r + 1, -val);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> Q;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    --v, u--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  a.push_back(BIT(n));
  mark[0] = 1;
  for (int i = 0; i < adj[0].size(); i++) {
    int u = adj[0][i], comp;
    lvl[u] = 1;
    comp = dfs(u, head);
    a.push_back(BIT(comp));
    ++head;
  }
  mark[0] = 0;
  for (int q = 0; q < Q; q++) {
    int t;
    cin >> t;
    if (t == 0) {
      int v, x, d, pos;
      cin >> v >> x >> d;
      --v;
      pos = mark[v];
      if (v == 0) {
        rootAns += x;
        addInterVal(0, 1, d, x);
      } else {
        addInterVal(pos, max(1, lvl[v] - d), min(a[pos].size, lvl[v] + d), x);
        if (d >= lvl[v]) {
          rootAns += x;
          addInterVal(0, 1, d - lvl[v], x);
          addInterVal(pos, 1, d - lvl[v], -x);
        }
      }
    } else {
      int v, pos;
      cin >> v;
      --v;
      pos = mark[v];
      if (v == 0)
        cout << rootAns;
      else
        cout << a[pos].get(lvl[v]) + a[0].get(lvl[v]);
      cout << endl;
    }
  }
  return 0;
}
