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;
int tree[2][1 << 17];
int dep[1 << 17];
vector<int> adj[1 << 17];
int nch, na, MX = 1 << 17;
int L[1 << 17], R[1 << 17], pos[1 << 17];
void dfs(int x, int p, int d) {
  dep[x] = d;
  pos[x] = na++;
  if (p == 1) L[nch] = pos[x];
  for (int i = 0; i < adj[x].size(); i++)
    if (adj[x][i] != p) dfs(adj[x][i], x, d + 1);
  R[nch] = na;
}
void add(int *fenwick, int x, int val) {
  if (!x) return;
  while (x < MX) {
    fenwick[x] += val;
    x += (x & -x);
  }
}
int sum(int *fenwick, int x) {
  int ret = 0;
  while (x) {
    ret += fenwick[x];
    x -= (x & -x);
  }
  return ret;
}
void Find(int v, int &l, int &r) {
  int low = 0, high = nch;
  while (low + 1 < high) {
    int mid = (low + high) / 2;
    if (L[mid] <= pos[v])
      low = mid;
    else
      high = mid;
  }
  l = L[low], r = R[low];
}
int main() {
  int n, q;
  int u, v, type, x, d;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  na = 2;
  pos[1] = 1;
  for (int i = 0; i < adj[1].size(); i++) dfs(adj[1][i], 1, 1), nch++;
  int root = 0;
  while (q--) {
    cin >> type;
    if (type) {
      cin >> v;
      if (v == 1)
        cout << root << endl;
      else {
        int ans = sum(tree[0], pos[v]);
        ans += sum(tree[1], dep[v]);
        cout << ans << endl;
      }
    } else {
      cin >> v >> x >> d;
      if (v == 1) {
        root += x;
        add(tree[1], 1, x);
        add(tree[1], d + 1, -x);
      } else {
        int rem = max(0, d - dep[v]);
        if (d >= dep[v]) root += x;
        if (rem) {
          add(tree[1], 1, x);
          add(tree[1], rem + 1, -x);
        }
        int mn, mx;
        Find(v, mn, mx);
        int l = max(mn + rem, pos[v] - d), r = min(mx, pos[v] + d + 1);
        if (l < r) {
          add(tree[0], l, x);
          add(tree[0], r, -x);
        }
      }
    }
  }
  return 0;
}
