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;
class tree {
 public:
  vector<int> s;
  void modify(int p, int v) {
    if (!p) return;
    for (; p < s.size(); p += ((p) & (-(p)))) s[p] += v;
  }
  int query(int p) {
    int ret = 0;
    for (; p; p -= ((p) & (-(p)))) ret += s[p];
    return ret;
  }
};
tree seg[110000], all;
int mapping[110000], belong[110000], tot, rootsum = 0, size[110000];
queue<int> q;
bool vis[110000];
class Graph {
 public:
  int head[110000], Next[110000 * 2], No[110000 * 2], tot;
  Graph() { tot = 1, memset(head, 0, sizeof head); }
  void add(int x, int y) {
    Next[tot] = head[x];
    head[x] = tot;
    No[tot++] = y;
  }
} G;
void bfs() {
  q.push(1);
  mapping[1] = 0;
  vis[1] = true;
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    vis[cur] = true;
    for (int i = G.head[cur]; i; i = G.Next[i]) {
      int y = G.No[i];
      if (vis[y]) continue;
      q.push(y);
      mapping[y] = mapping[cur] + 1;
      if (cur == 1)
        belong[y] = y;
      else
        belong[y] = belong[cur];
      size[belong[y]]++;
    }
  }
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    G.add(x, y), G.add(y, x);
  }
  all.s.resize(4 * n);
  bfs();
  for (int i = 2; i <= n; ++i) {
    if (belong[i] == i) seg[belong[i]].s.resize(size[belong[i]] * 8);
  }
  for (int i = 0; i < q; ++i) {
    int ty, v;
    scanf("%d%d", &ty, &v);
    if (!ty) {
      int x, d;
      scanf("%d%d", &x, &d);
      if (d < mapping[v]) {
        seg[belong[v]].modify(mapping[v] - d, x);
        seg[belong[v]].modify(min(mapping[v] + d + 1, size[belong[v]] + 1), -x);
      } else {
        rootsum += x;
        seg[belong[v]].modify(min(mapping[v] + d + 1, size[belong[v]] + 1), -x);
        all.modify(1, x);
        all.modify(d - mapping[v] + 1, -x);
        if (d - mapping[v] < mapping[v] + d) {
          seg[belong[v]].modify(d - mapping[v] + 1, x);
        }
      }
    } else {
      if (v == 1) {
        printf("%d\n", rootsum);
      } else {
        printf("%d\n",
               seg[belong[v]].query(mapping[v]) + all.query(mapping[v]));
      }
    }
  }
  return 0;
}
