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;
long long int n, q;
vector<long long int> adj[100005];
vector<long long int> tree[100005];
pair<long long int, long long int> ind[100005];
long long int root = 0;
void dfs(long long int i, long long int p, long long int in, long long int d) {
  ind[i] = make_pair(in, d);
  tree[in].push_back(0);
  long long int l = 1;
  for (auto j : adj[i]) {
    if (i == 1) {
      if (j != p) dfs(j, i, l++, d + 1);
    } else if (j != p)
      dfs(j, i, in, d + 1);
  }
}
long long int query(long long int I, long long int i) {
  long long int an = 0;
  while (i > 0) {
    an += tree[I][i];
    i -= (i & -i);
  }
  return an;
}
void upd(long long int I, long long int i, long long int v) {
  if (i <= 0) return;
  if (i >= tree[I].size()) i = tree[I].size() - 1;
  while (i < tree[I].size()) {
    tree[I][i] += v;
    i += (i & -i);
  }
}
int main() {
  scanf("%lld", &n);
  scanf("%lld", &q);
  for (int i = 1; i < n; i++) {
    long long int x, y;
    scanf("%lld", &x);
    scanf("%lld", &y);
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 1, 0, 0);
  for (int i = 1; i <= n; i++) tree[0].push_back(0), tree[i].push_back(0);
  for (int i = 1; i <= q; i++) {
    long long int c, v, x, d;
    scanf("%lld", &c);
    if (!c) {
      scanf("%lld", &v);
      scanf("%lld", &x);
      scanf("%lld", &d);
      if (v == 1) {
        upd(0, d, x);
        root += x;
        continue;
      }
      if (d >= ind[v].second) {
        upd(0, d - ind[v].second, x);
        root += x;
        upd(ind[v].first, d - ind[v].second, -x);
      }
      upd(ind[v].first, ind[v].second + d, x);
      upd(ind[v].first, ind[v].second - d - 1, -x);
    } else {
      scanf("%lld", &v);
      if (v == 1) {
        printf("%lld\n", root);
      } else {
        printf("%lld\n",
               query(0, n) - query(0, ind[v].second - 1) +
                   query(ind[v].first, tree[ind[v].first].size() - 1) -
                   query(ind[v].first, ind[v].second - 1));
      }
    }
  }
  return 0;
}
