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 = 100005;
int n, q, dis[N], root[N];
vector<int> g[N];
vector<vector<long long> > C;
int lowbit(int x) { return x & (-x); }
void Add(int st, int en, int val, int cur) {
  if (st > en) return;
  int sz = C[cur].size();
  for (int i = st; i < sz; i += lowbit(i)) C[cur][i] += val;
  for (int i = en + 1; i < sz; i += lowbit(i)) C[cur][i] -= val;
}
long long Sum(int deep, int cur) {
  long long ret = 0;
  int sz = C[cur].size();
  for (int i = deep; i > 0; i -= lowbit(i)) ret += C[cur][i];
  return ret;
}
void dfs(int u, int pre, int cnt, int cur) {
  dis[u] = cnt;
  root[u] = cur;
  C[cur].push_back(0);
  int sz = g[u].size();
  for (int i = 0; i < sz; i++) {
    int v = g[u][i];
    if (v != pre) dfs(v, u, cnt + 1, cur);
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i <= n; i++) g[i].clear();
  int a, b, op, l, r, x, v, d;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int sz = g[1].size();
  C.resize(sz + 5);
  for (int i = 0; i < sz; i++) {
    C[i + 1].push_back(0);
    dfs(g[1][i], 1, 1, i + 1);
  }
  C[0].resize(N, 0);
  long long sum = 0;
  while (q--) {
    scanf("%d", &op);
    if (op == 0) {
      scanf("%d%d%d", &v, &x, &d);
      int cur = root[v];
      if (d < dis[v])
        Add(dis[v] - d, dis[v] + d, x, cur);
      else {
        sum += x;
        Add(1, dis[v] + d, x, cur);
        Add(1, d - dis[v], x, 0);
        Add(1, d - dis[v], -x, cur);
      }
    } else {
      scanf("%d", &v);
      if (v == 1)
        printf("%I64d\n", sum);
      else {
        int cur = root[v];
        printf("%I64d\n", Sum(dis[v], cur) + Sum(dis[v], 0));
      }
    }
  }
  return 0;
}
