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 long N = 1e5 + 10;
vector<long long> adj[N];
vector<long long> chain[N];
long long c = 0;
long long ans[N];
void dfs(long long cur, long long par) {
  chain[c].push_back(cur);
  for (auto x : adj[cur]) {
    if (x == par) continue;
    dfs(x, cur);
  }
}
vector<vector<long long>> bit(N);
void add(long long index, long long val, long long ch) {
  for (long long i = index; i < (long long)bit[ch].size(); i = i | (i + 1))
    bit[ch][i] += val;
}
long long getSum(long long x, long long ch) {
  long long ans = 0;
  for (long long i = x; i >= 0; i = (i & (i + 1)) - 1) {
    ans += bit[ch][i];
  }
  return ans;
}
long long global[N];
void addG(long long index, long long val, long long n) {
  for (long long i = index; i < n; i = i | (i + 1)) global[i] += val;
}
long long getSumG(long long x) {
  long long ans = 0;
  for (long long i = x; i >= 0; i = (i & (i + 1)) - 1) {
    ans += global[i];
  }
  return ans;
}
int32_t main() {
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (auto x : adj[1]) {
    dfs(x, 1);
    c++;
  }
  map<long long, pair<long long, long long>> m;
  for (long long i = 0; i < c; i++) {
    for (long long j = 0; j < (long long)chain[i].size(); j++)
      m[chain[i][j]] = {i, j};
    bit[i].resize(chain[i].size() + 1);
  }
  while (q--) {
    long long type;
    cin >> type;
    if (type == 0) {
      long long v, x, d;
      cin >> v >> x >> d;
      if (v == 1) {
        ans[1] += x;
        addG(0, x, N);
        addG(d, -x, N);
      } else {
        long long ch = m[v].first;
        long long pos = m[v].second;
        if (d > pos + 1) {
          ans[1] += x;
          long long toAll = d - (pos + 1) - 1;
          addG(0, x, N);
          addG(toAll + 1, -x, N);
          add(toAll + 1, x, ch);
          add(pos + d + 1, -x, ch);
        } else if (d == pos + 1) {
          ans[1] += x;
          add(0, x, ch);
          add(pos + d + 1, -x, ch);
        } else {
          add(pos - d, x, ch);
          add(pos + d + 1, -x, ch);
        }
      }
    } else {
      long long v;
      cin >> v;
      if (v == 1) {
        cout << ans[1] << endl;
      } else {
        long long res = getSum(m[v].second, m[v].first);
        res += getSumG(m[v].second);
        cout << res << endl;
      }
    }
  }
  return 0;
}
