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;
struct node {
  int v;
  node() { ; }
  node(int x) { v = x; }
};
struct segtree {
  node* no;
  int n;
  inline int L(int i) { return i << 1; }
  inline int R(int i) { return i << 1 | 1; }
  segtree(int m) {
    n = 1;
    for (; n < m; n <<= 1)
      ;
    no = new node[n << 1];
    fill(no, no + n + n, node(0));
  }
  void update(int id, int l, int r, int ul, int ur, int x) {
    if (l >= ul && r <= ur) {
      no[id].v += x;
      return;
    }
    int m = (l + r) >> 1;
    if (l <= ur && m >= ul) update(L(id), l, m, ul, ur, x);
    if (m + 1 <= ur && r >= ul) update(R(id), m + 1, r, ul, ur, x);
  }
  void update(int l, int r, int x) { update(1, 1, n, l, min(r, n), x); }
  int query(int p) {
    p += n;
    int ret = 0;
    for (; p > 0; p >>= 1) ret += no[p].v;
    return ret;
  }
};
segtree* seg[100010];
vector<int> tr[100010];
int id[100010], dis[100010];
void dfs(int now, int p, int i, int d) {
  dis[now] = d;
  id[now] = i;
  for (int u = 0; u < tr[now].size(); u++) {
    if (tr[now][u] == p) continue;
    dfs(tr[now][u], now, i, d + 1);
  }
  if (tr[now].size() == 1) seg[i] = new segtree(d);
}
int main() {
  int n, q, tc;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    tr[x].push_back(y);
    tr[y].push_back(x);
  }
  dis[0] = 0;
  tc = tr[0].size();
  for (int i = 0; i < tr[0].size(); i++) {
    dfs(tr[0][i], 0, i, 1);
  }
  seg[tc] = new segtree(100010);
  for (int i = 0; i < q; i++) {
    int x, p, d;
    scanf("%d", &x);
    if (x == 0) {
      scanf("%d%d%d", &p, &x, &d);
      p--;
      if (p == 0)
        seg[tc]->update(1, d + 1, x);
      else if (d >= dis[p]) {
        seg[id[p]]->update(d - dis[p] + 1, d + dis[p], x);
        seg[tc]->update(1, d - dis[p] + 1, x);
      } else
        seg[id[p]]->update(dis[p] - d, dis[p] + d, x);
    } else {
      scanf("%d", &p);
      p--;
      int ret = 0;
      if (p == 0)
        ret = seg[tc]->query(0);
      else {
        ret = seg[id[p]]->query(dis[p] - 1);
        ret += seg[tc]->query(dis[p]);
      }
      printf("%d\n", ret);
    }
  }
  return 0;
}
