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 MAXN = 1e5;
struct BIT {
  vector<int> bit;
  int N;
  BIT(int N = 0) { resize(N); }
  void resize(int N) {
    this->N = N;
    bit.assign(N + 1, 0);
  }
  void update(int i, int v) {
    for (i++; i <= N; i += i & -i) bit[i] += v;
    return;
  }
  int query(int i) {
    int ret = 0;
    for (i++; i > 0; i -= i & -i) ret += bit[i];
    return ret;
  }
  void update(int s, int e, int v) {
    update(s, v);
    update(e + 1, -v);
  }
};
BIT branch[MAXN + 5];
BIT invRange;
int color[MAXN + 5], indx[MAXN + 5];
vector<vector<int> > G;
int n, qs;
int main() {
  int c, c2;
  scanf("%d%d", &n, &qs);
  G.resize(n);
  invRange.resize(n + 1);
  for (c = 0; c + 1 < n; c++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  indx[0] = 0;
  for (c = 0; c < G[0].size(); c++) {
    int p = 0, u = G[0][c];
    while (u != -1) {
      color[u] = c;
      indx[u] = indx[p] + 1;
      int v = -1;
      for (c2 = 0; c2 < G[u].size(); c2++)
        if (G[u][c2] != p) v = G[u][c2];
      p = u;
      u = v;
    }
    branch[c].resize(indx[p] + 1);
  }
  for (c = 0; c < qs; c++) {
    int type, u, x, d;
    scanf("%d%d", &type, &u);
    u--;
    int col = color[u], ix = indx[u];
    if (type == 0) {
      scanf("%d%d", &x, &d);
      int add = min(d, branch[col].N - ix - 1);
      d += add;
      ix += add;
      if (ix > d)
        branch[col].update(ix - d, ix, x);
      else {
        if (d - ix + 1 <= ix) branch[col].update(d - ix + 1, ix, x);
        invRange.update(n - (d - ix), x);
      }
    } else {
      int ret = invRange.query(n - ix);
      if (ix > 0) ret += branch[col].query(ix);
      printf("%d\n", ret);
    }
  }
  return 0;
}
