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 + 10;
struct segmentTree {
  int n = 0;
  vector<long long> t;
  vector<long long> lazy;
  void build(int len) {
    n = len + 1;
    t.resize(len * 4);
    lazy.resize(len * 4);
  }
  void push(int p, int l, int r) {
    if (!lazy[p]) return;
    if (l != r) lazy[p << 1] += lazy[p], lazy[p << 1 | 1] += lazy[p];
    t[p] += lazy[p] * (r - l + 1);
    lazy[p] = 0;
  }
  int ql, qr;
  void update(long long val) { update(val, 1, 0, n - 1); }
  void update(long long val, int p, int l, int r) {
    push(p, l, r);
    if (l > r || r < ql || l > qr) return;
    if (l >= ql && r <= qr) {
      lazy[p] += val;
      push(p, l, r);
      return;
    }
    update(val, p << 1, l, ((l + r) >> 1));
    update(val, p << 1 | 1, ((l + r) >> 1) + 1, r);
    t[p] = t[p << 1] + t[p << 1 | 1];
  }
  long long query() { return query(1, 0, n - 1); }
  long long query(int p, int l, int r) {
    push(p, l, r);
    if (l > r || r < ql || l > qr) return 0;
    if (l >= ql && r <= qr) return t[p];
    return query(p << 1, l, ((l + r) >> 1)) +
           query(p << 1 | 1, ((l + r) >> 1) + 1, r);
  }
} every_one;
vector<int> adj[N];
void dfs(vector<int>& p, int v, int par) {
  p.push_back(v);
  for (auto u : adj[v])
    if (u != par) {
      dfs(p, u, v);
    }
}
vector<int> branches[N];
long long root_num;
pair<int, int> node_place[N];
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  every_one.build(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--, y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  int branch_count = 0;
  for (auto x : adj[0]) {
    dfs(branches[branch_count++], x, 0);
  }
  for (int b = 0; b < branch_count; b++) {
    for (int i = 0; i < (int)branches[b].size(); i++) {
      node_place[branches[b][i]] = {b, i};
    }
  }
  vector<segmentTree> st(branch_count);
  for (int i = 0; i < branch_count; i++) {
    st[i].build(branches[i].size());
  }
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 0) {
      int v, x, d;
      scanf("%d %d %d", &v, &x, &d);
      v--;
      if (!v) {
        root_num += x;
        if (d) {
          every_one.ql = 0;
          every_one.qr = d - 1;
          every_one.update(x);
        }
        continue;
      }
      int b = node_place[v].first;
      int idx = node_place[v].second;
      int l = max(0, idx - d);
      int r = min(int(branches[b].size()) - 1, idx + d);
      st[b].ql = l, st[b].qr = r;
      st[b].update(x);
      if (idx < d) {
        root_num += x;
        if (idx + 1 < d) {
          int new_r = d - idx - 2;
          every_one.ql = 0;
          every_one.qr = new_r;
          every_one.update(x);
          st[b].ql = 0;
          st[b].qr = min(st[b].n - 1, new_r);
          st[b].update(-x);
        }
      }
    } else {
      int v;
      scanf("%d", &v);
      v--;
      if (!v) {
        printf("%I64d\n", root_num);
        continue;
      }
      int b = node_place[v].first;
      int idx = node_place[v].second;
      st[b].ql = st[b].qr = every_one.ql = every_one.qr = idx;
      long long ans = st[b].query();
      long long added = every_one.query();
      ans += added;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
