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;
struct BIT {
  vector<int> dat;
  void insert(int x, int v) {
    x = min(x, ((int)(dat).size()) - 1);
    while (x > 0) dat[x] += v, x -= x & -x;
  }
  int query(int x) {
    int ret = 0;
    while (x < ((int)(dat).size())) ret += dat[x], x += x & -x;
    return ret;
  }
} GL;
int END, cnt, n, m, grp[100010], dep[100010];
vector<BIT> V;
vector<int> G[100010];
void dfs(int p, int f, int c, int l) {
  grp[p] = c;
  dep[p] = l;
  V[c].dat.push_back(0);
  for (__typeof((G[p]).begin()) it = (G[p]).begin(); it != (G[p]).end(); it++)
    if (*it != f) dfs(*it, p, c, l + 1);
}
void INIT() {
  scanf("%d%d", &n, &m);
  GL.dat.push_back(0);
  GL.dat.push_back(0);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
    GL.dat.push_back(0);
  }
  cnt = 0;
  for (__typeof((G[1]).begin()) it = (G[1]).begin(); it != (G[1]).end(); it++) {
    V.push_back(BIT());
    V[cnt].dat.push_back(0);
    dfs(*it, 1, cnt++, 1);
  }
}
void QUERY() {
  int ty;
  scanf("%d", &ty);
  if (ty) {
    int x;
    scanf("%d", &x);
    if (x == 1)
      printf("%d\n", END);
    else
      printf("%d\n", GL.query(dep[x]) + V[grp[x]].query(dep[x]));
  } else {
    int v, x, d;
    scanf("%d%d%d", &v, &x, &d);
    if (v == 1) {
      END += x, GL.insert(d, x);
      return;
    }
    if (d >= dep[v]) {
      END += x;
      GL.insert(d - dep[v], x);
      V[grp[v]].insert(d - dep[v], -x);
    }
    V[grp[v]].insert(dep[v] + d, x);
    V[grp[v]].insert(dep[v] - d - 1, -x);
  }
}
int main() {
  INIT();
  while (m--) QUERY();
}
