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 = 110000;
const int inf = 1000000000;
int n, q, a, nr, b, niv[N], lant[N], size[N], re1;
vector<int> v[N], aint[N];
void df(int nod) {
  lant[nod] = nr;
  for (vector<int>::iterator it = v[nod].begin(); it != v[nod].end(); ++it)
    if (!lant[*it]) {
      niv[*it] = niv[nod] + 1;
      df(*it);
    }
}
void update(int arb, int nod, int pozx, int pozy, int poz1, int poz2, int nr) {
  if (pozx >= poz1 && poz2 >= pozy) {
    aint[arb][nod] += nr;
    return;
  }
  int mid = (pozx + pozy) / 2;
  if (mid >= poz1) update(arb, 2 * nod, pozx, mid, poz1, poz2, nr);
  if (mid < poz2) update(arb, 2 * nod + 1, mid + 1, pozy, poz1, poz2, nr);
}
int query(int arb, int nod, int pozx, int pozy, int poz) {
  if (pozx == pozy) return aint[arb][nod];
  int mid = (pozx + pozy) / 2;
  if (mid >= poz)
    return aint[arb][nod] + query(arb, 2 * nod, pozx, mid, poz);
  else
    return aint[arb][nod] + query(arb, 2 * nod + 1, mid + 1, pozy, poz);
}
int main() {
  int i, op, nod, nu, dist;
  cin >> n >> q;
  for (i = 1; i != n; ++i) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  lant[1] = -1;
  for (vector<int>::iterator it = v[1].begin(); it != v[1].end(); ++it) {
    ++nr;
    niv[*it] = 1;
    df(*it);
  }
  size[0] = n;
  for (i = 1; i <= n; ++i) size[lant[i]]++;
  for (i = 0; i <= nr; ++i) aint[i].resize(5 * (size[i] + 1));
  for (i = 1; i <= q; ++i) {
    cin >> op >> nod;
    if (op) {
      if (nod == 1) {
        cout << re1 << "\n";
        continue;
      }
      int a1 = query(0, 1, 1, size[0], niv[nod]);
      int a2 = 0;
      if (nod != 1) a2 = query(lant[nod], 1, 1, size[lant[nod]], niv[nod]);
      cout << a1 + a2 << "\n";
    } else {
      cin >> nu >> dist;
      if (nod != 1) {
        update(lant[nod], 1, 1, size[lant[nod]], max(1, niv[nod] - dist),
               min(niv[nod] + dist, size[lant[nod]]), nu);
        if (dist > niv[nod])
          update(lant[nod], 1, 1, size[lant[nod]], 1,
                 min(dist - niv[nod], size[lant[nod]]), -nu);
      }
      if (dist > niv[nod])
        update(0, 1, 1, size[0], 1, min(dist - niv[nod], size[0]), nu);
      if (dist >= niv[nod]) re1 += nu;
    }
  }
  return 0;
}
