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;
vector<vector<int> > e, arr, seg;
int num;
int seg_num[100001], ht[100001], seg_size[100001];
int cir_seg[5 * 100001];
int dfs(int current, int par, int h, int in) {
  arr[in].push_back(current);
  seg_num[current] = in;
  ht[current] = h;
  int sz = e[current].size();
  for (int i = 0; i < sz; i++) {
    if (e[current][i] != par) dfs(e[current][i], current, h + 1, in);
  }
  return 0;
}
int cir_update(int i, int l, int r, int x, int y, int v) {
  if (l > r || x > r || y < l) return 0;
  if (l >= x && r <= y) {
    cir_seg[i] += v;
    return 0;
  }
  cir_update((2 * i + 1), l, ((l + r) >> 1), x, y, v);
  cir_update((2 * i + 2), ((l + r) >> 1) + 1, r, x, y, v);
  return 0;
}
int cir_query(int i, int l, int r, int x) {
  if (l > r || x > r || x < l) return 0;
  if (l == r)
    return cir_seg[i];
  else
    return (cir_seg[i] + cir_query((2 * i + 1), l, ((l + r) >> 1), x) +
            cir_query((2 * i + 2), ((l + r) >> 1) + 1, r, x));
}
int update(int i, int l, int r, int x, int y, int v, int in) {
  if (l > r || x > r || y < l) return 0;
  if (l >= x && r <= y) {
    seg[in][i] += v;
    return 0;
  }
  update((2 * i + 1), l, ((l + r) >> 1), x, y, v, in);
  update((2 * i + 2), ((l + r) >> 1) + 1, r, x, y, v, in);
  return 0;
}
int query(int i, int l, int r, int x, int in) {
  if (l > r || x > r || x < l) return 0;
  if (l == r)
    return seg[in][i];
  else
    return (seg[in][i] + query((2 * i + 1), l, ((l + r) >> 1), x, in) +
            query((2 * i + 2), ((l + r) >> 1) + 1, r, x, in));
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  vector<int> temp;
  for (int i = 0; i <= n; i++) e.push_back(temp);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[x].push_back(y);
    e[y].push_back(x);
  }
  num = e[1].size();
  for (int i = 0; i < num; i++) {
    arr.push_back(temp);
    dfs(e[1][i], 1, 0, i);
    seg_size[i] = arr[i].size();
  }
  for (int i = 0; i < num; i++) {
    seg.push_back(temp);
    for (int j = 0; j < 5 * seg_size[i]; j++) seg[i].push_back(0);
  }
  while (q--) {
    int ch, v;
    scanf("%d%d", &ch, &v);
    if (ch) {
      if (v == 1)
        printf("%d\n", cir_query(0, 0, 100001 - 1, 0));
      else
        printf("%d\n",
               query(0, 0, seg_size[seg_num[v]] - 1, ht[v], seg_num[v]) +
                   cir_query(0, 0, 100001 - 1, ht[v] + 1));
    } else {
      int x, d;
      scanf("%d%d", &x, &d);
      if (v == 1) {
        cir_update(0, 0, 100001 - 1, 0, d, x);
      } else if (ht[v] - d >= 0) {
        update(0, 0, seg_size[seg_num[v]] - 1, ht[v] - d,
               min(ht[v] + d, seg_size[seg_num[v]] - 1), x, seg_num[v]);
      } else {
        int cir_range = d - ht[v] - 1;
        cir_update(0, 0, 100001 - 1, 0, cir_range, x);
        update(0, 0, seg_size[seg_num[v]] - 1, cir_range,
               min(ht[v] + d, seg_size[seg_num[v]] - 1), x, seg_num[v]);
      }
    }
  }
  return 0;
}
