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 = 2e5 + 10;
int part[N], h[N], n, m, v, u, cnt[N], t;
vector<int> g[N];
vector<long long> fen[2][N];
void add(bool typ, int c, int i, long long x) {
  for (i++; i > 0; i -= (-i) & i) fen[typ][c][i] += x;
}
long long get(bool typ, int c, int i) {
  long long res = 0;
  for (i++; i < fen[typ][c].size(); i += (-i) & i) res += fen[typ][c][i];
  return res;
}
void dfs(int v, int p, int t) {
  part[v] = t;
  h[v] = h[p] + 1;
  cnt[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    if (u != p) {
      int e = t;
      if (v == 0) e = i;
      dfs(u, v, e);
      cnt[v] += cnt[u];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i + 1 < n; i++) {
    cin >> v >> u;
    v--;
    u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  h[0] = -2;
  dfs(0, 0, 0);
  for (int j = 0; j < g[0].size(); j++) {
    int u = g[0][j];
    for (int i = 0; i <= 2 * cnt[u]; i++) fen[0][j].push_back(0);
  }
  for (int i = 0; i <= 2 * n; i++) fen[1][0].push_back(0);
  for (; m; m--) {
    int typ, x, d;
    cin >> typ >> v;
    v--;
    if (!typ) {
      cin >> x >> d;
      int l, r, l1;
      l = max(0, h[v] - d);
      r = min(cnt[g[0][part[v]]] - 1, h[v] + d);
      l1 = d - h[v];
      if (v == 0) {
        add(1, 0, d, x);
        continue;
      }
      add(0, part[v], r, x);
      if (l > 0) add(0, part[v], l - 1, -x);
      if (d > h[v]) {
        add(1, 0, l1 - 1, x);
        if (l1 > 1) add(0, part[v], min(cnt[g[0][part[v]]] - 1, l1 - 2), -x);
      }
    } else {
      if (v == 0)
        cout << get(1, 0, 0) << '\n';
      else
        cout << get(0, part[v], h[v]) + get(1, 0, h[v] + 1) << '\n';
    }
  }
}
