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;
vector<vector<int> > edges;
vector<int> head;
vector<int> dlen;
vector<int> dpos;
int dcount[100001];
int chno[100000];
bool vis[100000];
int depth[100000];
int dfs(int i, int d, int c) {
  vis[i] = 1;
  depth[i] = d;
  dcount[d]++;
  chno[i] = c;
  int m = 0;
  int k;
  for (int j = 0; j < edges[i].size(); j++)
    if (!vis[edges[i][j]]) {
      k = dfs(edges[i][j], d + 1, c);
      m = max(m, k);
    }
  return m + 1;
}
class fenwick {
 public:
  fenwick(int n) {
    size = n;
    T = new int[size + 1];
    for (int i = 0; i <= size; i++) T[i] = 0;
  }
  ~fenwick() {
    delete[] T;
    T = NULL;
  }
  void set(int i, int dif) {
    for (i++; i < size; i += (i & -i)) T[i] += dif;
  }
  int getSum(int i) {
    int c = 0;
    for (i++; i > 0; i -= (i & -i)) c += T[i];
    return c;
  }

 private:
  int *T;
  int size;
};
fenwick B(100005);
fenwick D(100005);
int main() {
  int N, Q;
  cin >> N >> Q;
  for (int i = 0; i < N; i++) edges.push_back(vector<int>());
  int a, b;
  for (int i = 1; i < N; i++) {
    cin >> a >> b;
    a--, b--;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  vis[0] = 1;
  depth[0] = 0;
  dcount[0]++;
  for (int j = 0; j < edges[0].size(); j++) {
    head.push_back(edges[0][j]);
    dlen.push_back(dfs(edges[0][j], 1, j));
    if (j == 0)
      dpos.push_back(0);
    else
      dpos.push_back(dpos[j - 1] + dlen[j - 1]);
  }
  for (int i = 1; i <= N; i++) dcount[i] += dcount[i - 1];
  int tp, v, x, d, low, high;
  for (int i = 0; i < Q; i++) {
    cin >> tp;
    if (tp == 0) {
      cin >> v >> x >> d;
      v--;
      if (v != 0) {
        if (d >= depth[v]) {
          b = d - depth[v];
          low = b + 1;
          high = min(dlen[chno[v]], depth[v] + d) + 1;
          B.set(0, +x);
          B.set(b + 1, -x);
          if (b < high) {
            D.set(dpos[chno[v]] - 1 + low, +x);
            D.set(dpos[chno[v]] - 1 + high, -x);
          }
        } else {
          low = depth[v] - d;
          high = min(dlen[chno[v]], depth[v] + d) + 1;
          D.set(dpos[chno[v]] - 1 + low, +x);
          D.set(dpos[chno[v]] - 1 + high, -x);
        }
      } else {
        B.set(0, +x);
        B.set(d + 1, -x);
      }
    } else {
      cin >> v;
      v--;
      if (v != 0)
        cout << D.getSum(dpos[chno[v]] - 1 + depth[v]) + B.getSum(depth[v])
             << endl;
      else
        cout << B.getSum(0) << endl;
    }
  }
}
