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> > V;
vector<int> adj[100005];
int common[100015];
int n;
int nBIT[100015], nBIT_sz[100015], nBIT_POS[100015];
void buildgraph() {
  int sz, i, par, cnt, now, sz1, temp;
  sz = adj[1].size();
  for (i = 0; i < sz; i++) {
    cnt = 0;
    par = 1;
    now = adj[1][i];
    nBIT[now] = i;
    nBIT_POS[now] = ++cnt;
    while (1) {
      sz1 = adj[now].size();
      if (sz1 == 1) break;
      temp = now;
      if (par == adj[now][0])
        now = adj[now][1];
      else
        now = adj[now][0];
      par = temp;
      nBIT[now] = i;
      nBIT_POS[now] = ++cnt;
    }
    nBIT_sz[i] = cnt;
    vector<int> Z(cnt + 1, 0);
    V.push_back(Z);
  }
}
int readIndiv(int tree_id, int idx) {
  int sum = 0;
  while (idx > 0) {
    sum += V[tree_id][idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void updateIndiv(int tree_id, int idx, int val) {
  while (idx <= nBIT_sz[tree_id]) {
    V[tree_id][idx] += val;
    idx += (idx & -idx);
  }
}
int readCommon(int idx) {
  int sum = 0;
  while (idx > 0) {
    sum += common[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void updateCommon(int idx, int val) {
  while (idx <= n + 10) {
    common[idx] += val;
    idx += (idx & -idx);
  }
}
int main() {
  int q, i, v, x, d, cmd;
  int u, len, ans;
  int branch_id, position, size_tree;
  scanf("%d%d", &n, &q);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  buildgraph();
  while (q--) {
    scanf("%d", &cmd);
    if (cmd == 0) {
      scanf("%d%d%d", &v, &x, &d);
      if (v == 1) {
        updateCommon(1, x);
        updateCommon(((d + 2) < (n + 1) ? (d + 2) : (n + 1)), -x);
      } else {
        branch_id = nBIT[v];
        position = nBIT_POS[v];
        size_tree = nBIT_sz[branch_id];
        updateIndiv(branch_id, ((1) > (position - d) ? (1) : (position - d)),
                    x);
        updateIndiv(branch_id,
                    ((position + d + 1) < (size_tree + 1) ? (position + d + 1)
                                                          : (size_tree + 1)),
                    -x);
        if (position - d <= 0) {
          len = d - position + 1;
          updateCommon(1, x);
          updateCommon(((len + 1) < (n + 1) ? (len + 1) : (n + 1)), -x);
          updateIndiv(branch_id, 1, -x);
          updateIndiv(branch_id,
                      ((len) < (size_tree + 1) ? (len) : (size_tree + 1)), x);
        }
      }
    } else {
      scanf("%d", &v);
      if (v == 1)
        ans = readCommon(1);
      else {
        branch_id = nBIT[v];
        position = nBIT_POS[v];
        size_tree = nBIT_sz[branch_id];
        ans = readCommon(position + 1);
        ans += readIndiv(branch_id, position);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
