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;
const int MAXN = 1e5 + 5;
int n, q;
int path[MAXN];
vector<int> sum[MAXN];
vector<int> adj[MAXN];
int dis[MAXN];
void dfs(int v, int fat) {
  int cnt = 0;
  bool b = false;
  for (int i = 0; i < (int)adj[v].size(); i++) {
    int u = adj[v][i];
    if (u == fat) continue;
    dis[u] = dis[v] + 1;
    if (v == 1) {
      cnt++;
      path[u] = cnt;
    } else
      path[u] = path[v];
    dfs(u, v);
    b = true;
  }
  if (v != 1 && !b) {
    sum[path[v]].resize(dis[v] + 2);
  }
}
void add(vector<int> &v, int x, int val) {
  if (x >= (int)v.size()) x = (int)v.size() - 1;
  for (; x > 0; x -= (x & -x)) v[x] += val;
}
int get(vector<int> &v, int x) {
  int t = 0;
  for (; x < (int)v.size(); x += (x & -x)) t += v[x];
  return t;
}
vector<int> sumroot;
int root = 0;
int main() {
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  sumroot.resize(n + 2);
  dfs(1, 0);
  for (int i = 0; i < q; i++) {
    int w;
    cin >> w;
    if (w == 0) {
      int v, x, d;
      cin >> v >> x >> d;
      if (v == 1) {
        add(sumroot, d, x);
        root += x;
      } else {
        add(sum[path[v]], dis[v] + d, x);
        if (dis[v] - d > 0)
          add(sum[path[v]], dis[v] - d - 1, -x);
        else {
          add(sumroot, d - dis[v], x);
          root += x;
          add(sum[path[v]], d - dis[v], -x);
        }
      }
    } else {
      int v;
      cin >> v;
      if (v == 1)
        cout << root << endl;
      else {
        int ans = get(sumroot, dis[v]);
        ans += get(sum[path[v]], dis[v]);
        cout << ans << endl;
      }
    }
  }
  return 0;
}
