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<int> adj[100001], tree[100001];
int n, b = 0, branch[100001], tam[100001], ord[100001], vis[100001];
void dfs(int v) {
  int i, u;
  vis[v] = 1;
  if (v == 1)
    ord[v] = 1;
  else
    ord[v] = ++tam[b];
  branch[v] = b;
  for (i = 0; i < adj[v].size(); i++) {
    u = adj[v][i];
    if (v == 1) {
      b = i + 1;
      tam[b] = 1;
    }
    if (!vis[u]) dfs(u);
  }
}
void update(int t, int p, int v) {
  for (; p; p -= p & -p) tree[t][p] += v;
}
void add(int v, int x, int d) {
  if (v == 1)
    update(0, d + 1, x);
  else {
    update(branch[v], min(tam[branch[v]], ord[v] + d), x);
    if (ord[v] - d > 1)
      update(branch[v], ord[v] - d - 1, -x);
    else {
      int m = d - ord[v] + 2;
      update(0, m, x);
      update(branch[v], min(tam[branch[v]], m), -x);
    }
  }
}
long long query(int t, int p) {
  long long ret = 0;
  for (; p <= tam[t]; p += p & -p) ret += tree[t][p];
  return ret;
}
long long value(int v) {
  if (v == 1) return query(0, 1);
  return query(0, ord[v]) + query(branch[v], ord[v]);
}
int main() {
  int q, u, v, i, j, c, x, d;
  scanf("%d%d", &n, &q);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (i = 1; i <= n; i++) vis[i] = 0;
  dfs(1);
  tam[0] = n;
  for (i = 0; i <= b; i++) {
    tree[i].resize(tam[i] + 1);
    for (j = 0; j <= tam[i]; j++) tree[i][j] = 0;
  }
  while (q--) {
    scanf("%d", &c);
    if (!c) {
      scanf("%d%d%d", &v, &x, &d);
      add(v, x, d);
    } else {
      scanf("%d", &v);
      cout << value(v) << endl;
    }
  }
  return 0;
}
