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 BIT {
 private:
  vector<long long> tree;
  int n;
  int LSOne(int x) { return x & (-x); }

 public:
  BIT(int x) {
    n = x;
    tree.resize(n + 1);
  }
  long long sum(int a) {
    a++;
    long long sum = 0;
    for (; a > 0; a -= LSOne(a)) sum += tree[a];
    return sum;
  }
  void update(int p, long long v) {
    for (; p < n + 1; p += LSOne(p)) tree[p] += v;
  }
  void addRange(int i, int j, long long v) {
    i++;
    j++;
    if (j < n) update(j + 1, -v);
    update(i, v);
  }
};
vector<vector<int> > adjList;
vector<int> pos, depth, branch, tin, tout;
int dfsPt = 1;
void dfs(int p, int parent, int b) {
  depth[p] = depth[parent] + 1;
  pos[p] = dfsPt++;
  branch[p] = b;
  for (int i : adjList[p])
    if (i != parent) dfs(i, p, b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  adjList.resize(n);
  pos.resize(n);
  depth.resize(n);
  branch.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a -= 1;
    b -= 1;
    adjList[a].push_back(b);
    adjList[b].push_back(a);
  }
  for (int i = 0; i < adjList[0].size(); i++) {
    branch[adjList[0][i]] = i;
    tin.push_back(dfsPt);
    dfs(adjList[0][i], 0, i);
    tout.push_back(dfsPt - 1);
  }
  BIT depthTree(*max_element(depth.begin(), depth.end()) + 1), dfsTree(dfsPt);
  for (int _ = 0; _ < q; _++) {
    int t;
    cin >> t;
    if (t == 0) {
      int p, v, d;
      cin >> p >> v >> d;
      p -= 1;
      int allD = d - depth[p];
      if (allD >= 0) depthTree.addRange(0, allD, v);
      if (p != 0) {
        int bs = tin[branch[p]], be = tout[branch[p]];
        if (allD < 0)
          dfsTree.addRange(max(bs, pos[p] - d), min(be, pos[p] + d), v);
        else if (bs + allD <= min(be, pos[p] + d))
          dfsTree.addRange(bs + allD, min(be, pos[p] + d), v);
      }
    } else {
      int p;
      cin >> p;
      p -= 1;
      cout << dfsTree.sum(pos[p]) + depthTree.sum(depth[p]) << "\n";
    }
  }
  return 0;
}
