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 double eps = 1.0e-11;
const double pi = acos(-1.0);
const int N = 100100;
vector<vector<int> > g;
int used[N];
int depth[N];
int color[N];
int colors[N];
void dfs(int v, int col, int dep) {
  used[v] = 1;
  color[v] = col;
  depth[v] = dep;
  for (int i = 0; i < (int)(g[v]).size(); i++) {
    int to = g[v][i];
    if (!used[to]) dfs(to, col, dep + 1);
  }
}
int global_fenwick[N];
void glob_add(int i, int x) {
  for (; i < N; i = (i | (i + 1))) global_fenwick[i] += x;
}
int glob_sum(int r) {
  int res = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) res += global_fenwick[r];
  return res;
}
int rglob_sum(int r) { return glob_sum(N - 1) - glob_sum(r - 1); }
vector<vector<int> > fenwick;
void add(int i, int x, int id) {
  for (; i < fenwick[id].size(); i = (i | (i + 1))) fenwick[id][i] += x;
}
int sum(int r, int id) {
  int res = 0;
  for (; r >= 0; r = (r & r + 1) - 1) res += fenwick[id][r];
  return res;
}
int rsum(int r, int id) {
  int sz = fenwick[id].size();
  return sum(sz - 1, id) - sum(r - 1, id);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  g.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  used[0] = 1;
  for (int i = 0; i < (int)(g[0]).size(); i++) {
    int to = g[0][i];
    if (!used[to]) {
      dfs(to, i, 1);
    }
  }
  for (int i = 1; i <= n - 1; i++) ++colors[color[i]];
  fenwick.resize(g[0].size());
  for (int i = 0; i < (int)(g[0]).size(); i++) {
    fenwick[i].resize(colors[i] + 1);
  }
  for (int query = 0; query < q; query++) {
    int type, v;
    scanf("%d%d", &type, &v);
    --v;
    int col = color[v];
    int dep = depth[v];
    if (type == 0) {
      int x, d;
      scanf("%d%d", &x, &d);
      if (dep > d) {
        add(dep - d - 1, -x, col);
        add(min(dep + d, (int)(fenwick[col]).size() - 1), x, col);
      } else {
        int glob_dep = d - dep;
        glob_add(glob_dep, x);
        add(min(glob_dep, (int)(fenwick[col]).size() - 1), -x, col);
        add(min(dep + d, (int)(fenwick[col]).size() - 1), x, col);
      }
    } else {
      int ans = rglob_sum(dep) + rsum(dep, col);
      printf("%d\n", ans);
    }
  }
  return 0;
}
