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;
long long al[100005], dep[100005], col[100005], n, a, b, c, d, q, val0;
vector<long long> fen[100005], v[100005];
void upd(long long x, long long v) {
  for (; x <= n; x += x & -x) al[x] += v;
}
long long que(long long x) {
  long long ret = 0;
  for (; x; x -= x & -x) ret += al[x];
  return ret;
}
void upd2(long long x, long long y, long long v) {
  if (x >= fen[y].size()) return;
  for (; x < fen[y].size(); x += x & -x) fen[y][x] += v;
}
long long que2(long long x, long long y) {
  long long ret = 0;
  for (; x; x -= x & -x) ret += fen[y][x];
  return ret;
}
void dfs(int x, int color, int y) {
  dep[x] = dep[y] + 1;
  col[x] = color;
  fen[color].push_back(0);
  for (int i = 0; i < v[x].size(); ++i)
    if (v[x][i] != y) dfs(v[x][i], color, x);
}
void print(long long x) {
  if (x == 1) return;
  cout << que(dep[x]) + que2(dep[x], col[x]) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 1; i <= n; ++i) fen[i].push_back(0);
  for (int i = 1; i <= v[1].size(); ++i) dfs(v[1][i - 1], i, 1);
  while (q--) {
    cin >> a;
    if (a == 0) {
      cin >> b >> c >> d;
      if (d > dep[b]) {
        int up = d - dep[b];
        upd(1, c);
        val0 += c;
        upd(up + 1, -c);
        upd2(up + 1, col[b], c);
        upd2(dep[b] + d + 1, col[b], -c);
      } else if (d <= dep[b]) {
        if (d == dep[b]) val0 += c;
        upd2(max(dep[b] - d, 1LL), col[b], c);
        upd2(dep[b] + d + 1, col[b], -c);
      }
    } else {
      cin >> b;
      if (b == 1)
        cout << val0 << '\n';
      else
        print(b);
    }
  }
}
