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 MOD = 1000000007;
const long long INF = std::numeric_limits<long long>::max();
const int MX = 100101;
const long double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270L;
template <typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) {
  return pair<T, T>(a.first + b.first, a.second + b.second);
}
template <typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) {
  return pair<T, T>(a.first - b.first, a.second - b.second);
}
template <typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) {
  return (a.first * b.first + a.second * b.second);
}
template <typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) {
  return (a.first * b.second - a.second * b.first);
}
vector<int> edges[MX];
int N;
vector<int> dep;
vector<int> kur;
vector<int> sizes;
int bs = 0;
void dfs(int first, int p, int b) {
  sizes[b]++;
  kur[first] = b;
  dep[first] = dep[p] + 1;
  for (auto u : edges[first])
    if (u != p) dfs(u, first, b);
}
struct FENWICK {
  vector<int> A;
  int sz;
  FENWICK(int n) : sz(n) { A.resize(n + 1, 0); }
  void add(int i, int first) {
    for (; i <= sz; i += (i) & (-i)) A[i] += first;
  }
  void add(int i, int j, int first) {
    add(i, first);
    if (j + 1 <= sz) add(j + 1, -first);
  }
  int getQ(int i) {
    int get = 0;
    for (; i > 0; i -= (i) & (-i)) get += A[i];
    return get;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int Q;
  cin >> N >> Q;
  dep = vector<int>(N + 1);
  kur = vector<int>(N + 1);
  int a, b;
  for (int i = 0; i < N - 1; ++i) {
    cin >> a >> b;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  dep[1] = 1;
  for (auto u : edges[1]) sizes.push_back(1), dfs(u, 1, bs++);
  vector<FENWICK> mazi;
  int mxsize = *max_element(sizes.begin(), sizes.end());
  FENWICK didelis(mxsize);
  for (int i = 0; i < bs; ++i) mazi.push_back(FENWICK(sizes[i]));
  for (int i = 0; i < Q; ++i) {
    int t;
    cin >> t;
    if (!t) {
      int v, first, d;
      cin >> v >> first >> d;
      if (v == 1) {
        didelis.add(1, min(mxsize, d + 1), first);
      } else {
        int kuris = kur[v];
        int depth = dep[v];
        int prieroot = max(0, d - (depth - 1) + 1);
        if (prieroot) didelis.add(1, min(prieroot, mxsize), first);
        int nuo = max(prieroot + 1, depth - d);
        int iki = min(sizes[kuris], depth + d);
        if (nuo <= iki) mazi[kuris].add(nuo, iki, first);
      }
    } else {
      int v;
      cin >> v;
      int ret = 0;
      if (v == 1) {
        printf("%d\n", didelis.getQ(1));
        continue;
      }
      ret += didelis.getQ(dep[v]);
      ret += mazi[kur[v]].getQ(dep[v]);
      printf("%d\n", ret);
    }
  }
}
