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 = 1000 * 100 + 10;
int d[MAXN], ind[MAXN];
vector<int> ng[MAXN];
vector<int> path[MAXN];
struct Seg {
  Seg() {}
  void set(int sizeT) {
    size = sizeT;
    a.resize(4 * size, 0);
  }
  void shift(int cur) {
    a[cur * 2] += a[cur];
    a[cur * 2 + 1] += a[cur];
    a[cur] = 0;
  }
  void add(int lo, int hi, int val, int cur, int s, int e) {
    if (lo >= hi) return;
    if (lo == s && hi == e) {
      a[cur] += val;
      return;
    }
    shift(cur);
    int mid = (s + e) / 2;
    if (lo < mid) add(lo, min(mid, hi), val, cur * 2, s, mid);
    if (hi > mid) add(max(lo, mid), hi, val, cur * 2 + 1, mid, e);
  }
  int ask(int plc, int cur, int s, int e) {
    if (e - s < 2) return a[cur];
    shift(cur);
    int mid = (s + e) / 2;
    if (plc < mid)
      return ask(plc, cur * 2, s, mid);
    else
      return ask(plc, cur * 2 + 1, mid, e);
  }
  vector<int> a;
  int size;
} seg[MAXN];
void dfs(int x, int par, int id, int dep) {
  path[id].push_back(x);
  ind[x] = id;
  d[x] = dep;
  for (int i = 0; i < ng[x].size(); i++) {
    int u = ng[x][i];
    if (u == par) continue;
    dfs(u, x, id, dep + 1);
  }
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    ng[a].push_back(b);
    ng[b].push_back(a);
  }
  for (int i = 0; i < ng[0].size(); i++) {
    int u = ng[0][i];
    dfs(u, 0, i + 1, 1);
    seg[i + 1].set((int)path[i + 1].size());
  }
  seg[0].set(n);
  for (int i = 0; i < q; i++) {
    int t, v, add, dist;
    cin >> t >> v;
    v--;
    if (!t) {
      cin >> add >> dist;
      if (v == 0) {
        seg[0].add(0, dist + 1, add, 1, 0, n);
        continue;
      }
      int st = d[v] - dist - 1, end = min(d[v] + dist, seg[ind[v]].size);
      if (dist >= d[v]) {
        seg[0].add(0, dist - d[v] + 1, add, 1, 0, n);
        st = dist - d[v];
      }
      seg[ind[v]].add(st, end, add, 1, 0, seg[ind[v]].size);
    } else {
      int ans = seg[0].ask(d[v], 1, 0, n);
      if (v != 0) ans += seg[ind[v]].ask(d[v] - 1, 1, 0, seg[ind[v]].size);
      cout << ans << endl;
    }
  }
  return 0;
}
