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 = 100 * 1000 + 7;
int n, q;
struct Fen {
  vector<long long> fen;
  void init(int n) {
    for (int i = 0; i < n + 2; ++i) fen.push_back(0);
  }
  void add(int idx, long long val) {
    for (; idx < (long long)(fen).size(); idx += idx & -idx) {
      fen[idx] += val;
    }
  }
  long long get(int idx) {
    idx--;
    long long sum = 0;
    for (; idx; idx -= idx & -idx) {
      sum += fen[idx];
    }
    return sum;
  }
};
vector<int> adj[MAXN], curr[MAXN];
bool isPassed[MAXN];
int brnchCnt, brnch[MAXN], sPos[MAXN], ePos[MAXN], zerVal;
long long maxBrnch;
void dfs(int root) {
  isPassed[root] = true;
  for (auto nei : adj[root]) {
    if (!isPassed[nei]) {
      root == 1 ? brnch[nei] = ++brnchCnt : brnch[nei] = brnch[root];
      curr[brnch[nei]].push_back(nei);
      sPos[nei] = (long long)(curr[brnch[nei]]).size();
      maxBrnch = max(maxBrnch, (long long)(curr[brnch[nei]]).size());
      dfs(nei);
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i < n; ++i) {
    int tmp1, tmp2;
    cin >> tmp1 >> tmp2;
    adj[tmp1].push_back(tmp2);
    adj[tmp2].push_back(tmp1);
  }
  dfs(1);
  Fen pAns[brnchCnt + 7];
  pAns[0].init(maxBrnch + 7);
  for (int i = 1; i < brnchCnt + 1; ++i)
    pAns[i].init((long long)(curr[i]).size() + 7);
  while (q--) {
    int type;
    cin >> type;
    if (type) {
      int v;
      cin >> v;
      if (v == 1) {
        cout << zerVal << endl;
      } else
        cout << (pAns[0].get(sPos[v] + 2) + pAns[brnch[v]].get(sPos[v] + 1))
             << endl;
    } else {
      long long v, x, d;
      cin >> v >> x >> d;
      if (v == 1) {
        zerVal += x;
        pAns[0].add(1, x);
        pAns[0].add(d + 2, -x);
      } else {
        long long tmp2 =
            min(sPos[v] + d + 1, (long long)(pAns[brnch[v]].fen).size() + 1);
        if (d < sPos[v]) {
          pAns[brnch[v]].add(sPos[v] - d, x);
          pAns[brnch[v]].add(tmp2, -x);
        } else {
          pAns[0].add(1, x);
          pAns[0].add(d - sPos[v] + 2, -x);
          zerVal += x;
          pAns[brnch[v]].add(d - sPos[v] + 1, x);
          pAns[brnch[v]].add(tmp2, -x);
        }
      }
    }
  }
  return 0;
}
