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;
int fen[2][MAXN], st[MAXN], en[MAXN], now = 0, h[MAXN];
vector<int> g[MAXN];
int n, q, H;
void dfs(int v, int hei, int p = 0) {
  H = max(H, hei);
  h[v] = hei;
  st[v] = now++;
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    if (u != p) dfs(u, hei + 1, v);
  }
  en[v] = now;
}
void add(int x, int val, int id) {
  x++;
  for (; x <= n; x += x & -x) fen[id][x] += val;
}
int get(int x, int id) {
  x++;
  int ret = 0;
  for (; x; x -= x & -x) ret += fen[id][x];
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0, 0);
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (!t) {
      int v, x, d;
      cin >> v >> x >> d;
      v--;
      if (v == 0) {
        add(0, x, 1);
        add(min(H, d) + 1, -x, 1);
        continue;
      }
      if (h[v] > d) {
        add(st[v] - d, x, 0);
        add(min(st[v] + d + 1, en[v]), -x, 0);
      } else {
        int ha = min(d - h[v], H);
        add(0, x, 1);
        add(ha + 1, -x, 1);
        int s = min(en[v], st[v] - (h[v] - ha) + 1),
            e = min(st[v] + d + 1, en[v]);
        add(s, x, 0);
        add(e, -x, 0);
      }
    } else {
      int v;
      cin >> v;
      v--;
      cout << get(st[v], 0) + get(h[v], 1) << endl;
    }
  }
  return 0;
}
