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 N = 1e5 + 5;
int s[N], e[N], sum[N * 2], t[N * 2], d[N], ind[N], pos = 1, par[N];
vector<int> adj[N];
void dfs(int x, int p) {
  if (p == 1) par[x] = x;
  if (p > 1) par[x] = par[p];
  ind[x] = pos;
  s[x] = pos++;
  int u, i;
  d[x] = d[p] + 1;
  for (i = 0; i < adj[x].size(); ++i) {
    u = adj[x][i];
    if (u != p) dfs(u, x);
  }
  e[x] = pos - 1;
}
void su(int l, int r, int v, int ty) {
  int *p = t;
  if (ty) p = sum;
  for (l += N, r += N; l < r; l >>= 1, r >>= 1) {
    if (l & 1) p[l++] += v;
    if (r & 1) p[--r] += v;
  }
}
int sq(int l, int ty) {
  int ret = 0;
  l += N;
  int *p = t;
  if (ty) p = sum;
  for (; l > 0; l >>= 1) ret += p[l];
  return ret;
}
int main() {
  int type, v, x, y, n, i, a, b, q, st, lt, p;
  cin >> n >> q;
  for (i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  d[0] = -1;
  dfs(1, 0);
  while (q--) {
    scanf("%d", &type);
    if (type == 0) {
      scanf("%d%d%d", &v, &x, &y);
      if (y >= d[v]) {
        a = y - d[v];
        su(0, a + 1, x, 1);
        p = par[v];
        lt = e[p], st = ind[p] + a;
        lt = min(ind[v] + y, lt);
      } else {
        p = par[v];
        lt = e[p];
        st = ind[p];
        st = max(st, ind[v] - y);
        lt = min(lt, ind[v] + y);
      }
      if (v != 1) su(st - 1, lt, x, 0);
    } else {
      scanf("%d", &x);
      printf("%d\n", sq(d[x], 1) + sq(ind[x] - 1, 0));
    }
  }
}
