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<int> g[100010];
vector<int> B[100010];
int depth[100010];
int num[100010];
void dfs(int x, int parent, int level, int idx) {
  B[idx].push_back(0);
  depth[x] = level;
  num[x] = idx;
  for (int i = 0; i < g[x].size(); i++) {
    if (g[x][i] != parent) {
      dfs(g[x][i], x, level + 1, idx);
    }
  }
}
void update(int v, int x, int idx) {
  for (int i = v; i < B[idx].size(); i += (i & (-i))) B[idx][i] += x;
}
int query(int v, int idx) {
  int ans = 0;
  for (int i = v; i > 0; i -= (i & (-i))) ans += B[idx][i];
  return ans;
}
int main() {
  int n, m, q;
  cin >> n >> q;
  m = n - 1;
  while (m--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 0; i < g[0].size(); i++) {
    B[i].clear();
    B[i].push_back(0);
    dfs(g[0][i], 0, 1, i);
  }
  for (int i = 0; i < n + 5; i++) {
    B[g[0].size()].push_back(0);
  }
  int root = 0;
  while (q--) {
    int type;
    cin >> type;
    if (type == 0) {
      int v, x, dist;
      cin >> v >> x >> dist;
      v--;
      if (v == 0) {
        root += x;
        update(1, x, g[0].size());
        update(dist + 1, -x, g[0].size());
      } else {
        if (dist >= depth[v]) {
          int rem = dist - depth[v];
          root += x;
          update(1, x, g[0].size());
          update(rem + 1, -x, g[0].size());
          update(rem + 1, x, num[v]);
          update(depth[v] + dist + 1, -x, num[v]);
        } else {
          update(depth[v] - dist, x, num[v]);
          update(depth[v] + dist + 1, -x, num[v]);
        }
      }
    }
    if (type == 1) {
      int v;
      cin >> v;
      v--;
      if (v == 0)
        cout << root << endl;
      else {
        int ans = query(depth[v], num[v]) + query(depth[v], g[0].size());
        cout << ans << endl;
      }
    }
  }
  return 0;
}
