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 + 10;
int fn[2][MAXN], st[MAXN], fin[MAXN], d[MAXN], ti;
vector<int> a[MAXN];
void dfs(int v, int par) {
  st[v] = ti++;
  for (auto u : a[v]) {
    if (u == par) continue;
    d[u] = d[v] + 1;
    dfs(u, v);
  }
  fin[v] = ti;
}
void add(int p, int val, int id) {
  for (++p; p < MAXN; p += p & -p) fn[id][p] += val;
}
int get(int p, int id) {
  int res = 0;
  for (; p > 0; p -= p & -p) res += fn[id][p];
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 0) {
      int v, x, dis;
      cin >> v >> x >> dis;
      if (v == 1) {
        add(dis, x, 1);
        continue;
      }
      int down = min(fin[v], st[v] + dis + 1), up;
      if (dis < d[v])
        up = st[v] - dis;
      else {
        int dis2 = dis - d[v];
        dis = 2 * d[v] - dis - 1;
        up = st[v] - dis;
        add(dis2, x, 1);
      }
      if (up < down) {
        add(up, x, 0);
        add(down, -x, 0);
      }
    } else {
      int v;
      cin >> v;
      int ans = get(st[v] + 1, 0) + get(n, 1) - get(d[v], 1);
      cout << ans << endl;
    }
  }
  return 0;
}
