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 {
  int n;
  vector<int> ary;
  BIT(int size) : n(size), ary(size + 1 + 1) {}
  void insert(int x, int v) {
    if (x < 0) return;
    x = min(n, x);
    for (int i = x + 1; i > 0; i -= i & (-i)) ary[i] += v;
  }
  void insert(int l, int r, int v) {
    if (l > r) return;
    insert(l - 1, -v);
    insert(r, v);
  }
  int query(int x) {
    x = min(n, x);
    int s = 0;
    for (int i = x + 1; i <= n + 1; i += i & (-i)) s += ary[i];
    return s;
  }
};
vector<int> e[100010];
vector<int> chain[100010];
pair<int, int> table_c[100010];
BIT* bits[100010];
bool vis[100010];
BIT* prefix;
void dfs(int now, int counter) {
  if (vis[now]) return;
  vis[now] = 1;
  chain[counter].push_back(now);
  for (size_t i = 0; i < e[now].size(); i++) {
    int b = e[now][i];
    if (!vis[b]) dfs(b, counter);
  }
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    --a, --b;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  memset(vis, 0, sizeof(vis));
  vis[0] = 1;
  int counter = 0;
  for (size_t i = 0; i < e[0].size(); i++) {
    int b = e[0][i];
    dfs(b, counter);
    counter++;
  }
  for (int i = 0; i < counter; i++) {
    for (size_t j = 0; j < chain[i].size(); j++) {
      int v = chain[i][j];
      table_c[v] = make_pair(i, j);
    }
    bits[i] = new BIT(chain[i].size());
  }
  prefix = new BIT(n);
  while (q--) {
    int op, v;
    scanf("%d %d", &op, &v);
    --v;
    pair<int, int> p = table_c[v];
    if (!op) {
      int x, d;
      scanf("%d %d", &x, &d);
      if (v == 0) {
        prefix->insert(0, d, x);
      } else {
        int l = p.second - d;
        int r = p.second + d;
        l = max(l, 0);
        bits[p.first]->insert(l, r, x);
        r = d - (p.second + 1);
        prefix->insert(0, r, x);
        bits[p.first]->insert(0, r - 1, -x);
      }
    } else {
      int ans = 0;
      if (v == 0)
        ans = prefix->query(0);
      else {
        int a = bits[p.first]->query(p.second);
        int b = prefix->query(p.second + 1);
        ans = a + b;
      }
      printf("%d\n", ans);
    }
  }
}
