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;
int n, q, a, b, v, d, x, t, mx;
int l[200005];
int r[200005];
int z[200005];
vector<vector<int> > lbt;
vector<int> e[200005];
void trv(int k, int f) {
  l[k] = l[f] + 1;
  z[r[k]] = l[k];
  for (int i = 0, br = r[k]; i < e[k].size(); i++) {
    if (e[k][i] != f) {
      r[e[k][i]] = br;
      br++;
      trv(e[k][i], k);
    }
  }
}
void upbt(int k, int a, int b, int x) {
  for (int i = a; i < lbt[k].size(); i += i & (-i)) {
    lbt[k][i] += x;
  }
  for (int i = b + 1; i < lbt[k].size(); i += i & (-i)) {
    lbt[k][i] -= x;
  }
}
int qubt(int k, int a) {
  int s = 0;
  for (int i = a; i > 0; i -= i & (-i)) {
    s += lbt[k][i];
  }
  return s;
}
void up(int k, int x, int d) {
  if (k == 0) {
    upbt(e[0].size(), 1, d + 1, x);
    return;
  }
  if (1 < l[k] - d) {
    upbt(r[k], l[k] - d, min(z[r[k]], l[k] + d), x);
  } else {
    upbt(r[k], 2, min(z[r[k]], l[k] + d), x);
    upbt(e[0].size(), 1, d - l[k] + 2, x);
    if (d - l[k] + 2 > 1) {
      upbt(r[k], 2, min(z[r[k]], d - l[k] + 2), -x);
    }
  }
}
int qu(int k) {
  int ret = qubt(e[0].size(), l[k]);
  if (k != 0) {
    ret += qubt(r[k], l[k]);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--;
    b--;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  trv(0, 0);
  for (int i = 0; i < e[0].size(); i++) {
    lbt.push_back(vector<int>(z[i] + 5));
    mx = max(mx, z[i]);
  }
  lbt.push_back(vector<int>(mx + 5));
  for (int i = 0; i < q; i++) {
    cin >> t >> v;
    v--;
    if (t == 0) {
      cin >> x >> d;
      up(v, x, d);
    } else {
      cout << qu(v) << "\n";
    }
  }
  return 0;
}
