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 + 5;
int n, q, x, y, root, vertex, d, colorCnt, type, depth[maxn], chain[maxn],
    length[maxn];
vector<int> adj[maxn];
vector<vector<int> > BIT;
void dfs(int curr, int prev, int col) {
  depth[curr] = depth[prev] + 1;
  chain[curr] = col;
  length[col]++;
  for (vector<int>::iterator it = adj[curr].begin(); it != adj[curr].end();
       it++)
    if (*it != prev) dfs(*it, curr, col);
}
void update(int col, int x, int val) {
  while (x <= length[col]) {
    BIT[col][x] += val;
    x += x & -x;
  }
}
int get(int col, int x) {
  int ret = 0;
  while (x) {
    ret += BIT[col][x];
    x -= x & -x;
  }
  return ret;
}
void addRoot(int x, int d, int except) {
  root += x;
  update(colorCnt, 1, x);
  update(colorCnt, min(n, d + 1), -x);
  if (except == -1) return;
  update(chain[except], 1, -x);
  update(chain[except], min(length[chain[except]] + 1, d + 1), x);
}
void add() {
  scanf("%d %d %d", &vertex, &x, &d);
  vertex--;
  if (!vertex)
    addRoot(x, d, -1);
  else {
    update(chain[vertex], max(1, depth[vertex] - d), x);
    update(chain[vertex], min(length[chain[vertex]] + 1, depth[vertex] + d + 1),
           -x);
    if (depth[vertex] - d <= 0) addRoot(x, d - depth[vertex], vertex);
  }
}
void solve() {
  scanf("%d", &vertex);
  vertex--;
  if (!vertex)
    printf("%d\n", root);
  else
    printf("%d\n",
           get(chain[vertex], depth[vertex]) + get(colorCnt, depth[vertex]));
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &x, &y);
    adj[x - 1].push_back(y - 1);
    adj[y - 1].push_back(x - 1);
  }
  for (vector<int>::iterator it = adj[0].begin(); it != adj[0].end(); it++)
    dfs(*it, 0, colorCnt++);
  BIT.resize(colorCnt + 5);
  for (int i = 0; i < colorCnt; i++) BIT[i].resize(length[i] + 5);
  BIT[colorCnt].resize(n + 5);
  length[colorCnt] = n;
  while (q--) {
    scanf("%d", &type);
    if (!type)
      add();
    else
      solve();
  }
  return 0;
}
