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;
void upd(int idx, int v, int lim, int *tree) {
  while (idx <= lim) {
    tree[idx] += v;
    idx += idx & -idx;
  }
}
void upd(int idx, int v, int lim, vector<int> &tree) {
  while (idx <= lim) {
    tree[idx] += v;
    idx += idx & -idx;
  }
}
int get(int idx, int *tree) {
  int ret = 0;
  while (idx > 0) {
    ret += tree[idx];
    idx -= idx & -idx;
  }
  return ret;
}
int get(int idx, vector<int> &tree) {
  int ret = 0;
  while (idx > 0) {
    ret += tree[idx];
    idx -= idx & -idx;
  }
  return ret;
}
vector<int> edge[200000], tree[200000];
int cir[200000];
int under[200000], num[200000];
void call(int u, int p) {
  tree[under[p]].push_back(0);
  num[u] = num[p] + 1;
  under[u] = under[p];
  for (__typeof(((int)edge[u].size())) i = 0; i < (((int)edge[u].size()));
       i++) {
    int v = edge[u][i];
    if (v == p) continue;
    call(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, q;
  cin >> n >> q;
  for (__typeof(n - 1) i = 0; i < (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    edge[u - 1].push_back(v - 1);
    edge[v - 1].push_back(u - 1);
  }
  for (__typeof(((int)edge[0].size())) i = 0; i < (((int)edge[0].size()));
       i++) {
    int v = edge[0][i];
    under[0] = v;
    tree[v].push_back(0);
    call(v, 0);
  }
  while (q--) {
    int type, v, x, d;
    cin >> type >> v;
    v--;
    if (!type) {
      cin >> x >> d;
      if (d >= num[v]) {
        upd(1, x, n, cir);
        upd(d - num[v] + 2, -x, n, cir);
        int idx = under[v];
        int from = d - num[v] + 1;
        int to = min(num[v] + d, ((int)tree[idx].size()) - 1);
        upd(from, x, ((int)tree[idx].size()) - 1, tree[idx]);
        upd(to + 1, -x, ((int)tree[idx].size()) - 1, tree[idx]);
      } else {
        int idx = under[v];
        int from = num[v] - d;
        int to = min(num[v] + d, ((int)tree[idx].size()) - 1);
        upd(from, x, ((int)tree[idx].size()) - 1, tree[idx]);
        upd(to + 1, -x, ((int)tree[idx].size()) - 1, tree[idx]);
      }
    } else {
      int ans = get(num[v] + 1, cir);
      if (v) {
        int idx = under[v];
        ans += get(num[v], tree[idx]);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
