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 long int mod = 1e9 + 7;
long int n, q;
vector<long int> adj[100005];
vector<vector<long int> > ch, chain;
pair<long int, long int> chainID[100005];
long int mxChainSize, ans;
vector<vector<long int> > seg;
void update(long int id, long int idx, long int l, long int r, long int &frm,
            long int &to, long int &val) {
  if (l > to || r < frm) return;
  if (l >= frm && r <= to) {
    seg[id][idx] += val;
    return;
  }
  update(id, 2 * idx, l, (l + r >> 1), frm, to, val);
  update(id, 2 * idx + 1, (l + r >> 1) + 1, r, frm, to, val);
  return;
}
void query(long int id, long int idx, long int l, long int r, long int &num) {
  if (l > num || r < num) return;
  ans += seg[id][idx];
  if (l == num && r == num) return;
  query(id, 2 * idx, l, (l + r >> 1), num);
  query(id, 2 * idx + 1, (l + r >> 1) + 1, r, num);
  return;
}
void dfs() {
  stack<pair<pair<long int, long int>, long int> > stk;
  while (!stk.empty()) stk.pop();
  stk.push(make_pair(make_pair(1, 0), 0));
  ch.resize(adj[1].size() + 1);
  while (!stk.empty()) {
    long int node = ((stk.top()).first).first;
    long int par = ((stk.top()).first).second;
    long int id = (stk.top()).second;
    stk.pop();
    for (long int i = 0; i < adj[node].size(); i++) {
      long int nxt = adj[node][i];
      if (nxt == par) continue;
      if (node == 1) ch[++id].push_back(node);
      ch[id].push_back(nxt);
      stk.push(make_pair(make_pair(nxt, node), id));
    }
  }
  vector<pair<long int, long int> > srt(0);
  for (long int i = 1; i < ch.size(); i++)
    srt.push_back(make_pair(ch[i].size(), i));
  sort(srt.begin(), srt.end());
  for (long int i = 0; i < srt.size(); i++) chain.push_back(ch[srt[i].second]);
  for (long int i = 0; i < chain.size(); i++)
    for (long int j = 0; j < chain[i].size(); j++)
      chainID[chain[i][j]] = make_pair(i, j + 1);
  mxChainSize = chain[chain.size() - 1].size();
  return;
}
void solve() {
  scanf("%ld %ld", &n, &q);
  for (long int i = 0; i < n - 1; i++) {
    long int u, v;
    scanf("%ld %ld", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs();
  seg.resize(chain.size() + 1);
  for (long int i = 0; i < chain.size(); i++)
    seg[i].resize(4 * chain[i].size());
  seg[chain.size()].resize(4 * mxChainSize);
  long int one = 0;
  while (q--) {
    long int typ;
    scanf("%ld", &typ);
    if (typ == 0) {
      long int v, x, d, frm, to, val;
      scanf("%ld %ld %ld", &v, &x, &d);
      if (v == 1) one += x;
      if (v != 1) {
        update(chainID[v].first, 1, 1, chain[chainID[v].first].size(),
               frm = max((long int)1, chainID[v].second - d),
               to = min(chainID[v].second + d,
                        (long int)chain[chainID[v].first].size()),
               x);
        if (d >= chainID[v].second)
          update(chainID[v].first, 1, 1, chain[chainID[v].first].size(),
                 frm = 1, to = d - chainID[v].second + 2, val = -x);
        if (d >= chainID[v].second - 1) one += x;
      }
      if (d >= chainID[v].second)
        update(chain.size(), 1, 1, mxChainSize, frm = 1,
               to = d - chainID[v].second + 2, x);
    }
    if (typ == 1) {
      long int v;
      scanf("%ld", &v);
      if (v == 1) {
        printf("%ld\n", one);
        continue;
      }
      ans = 0;
      query(chain.size(), 1, 1, mxChainSize, chainID[v].second);
      query(chainID[v].first, 1, 1, chain[chainID[v].first].size(),
            chainID[v].second);
      printf("%ld\n", ans);
    }
  }
  return;
}
int main() {
  solve();
  return 0;
}
