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> a[200003];
int dem = 0, n, m;
typedef int mang[200003];
int c[2][200003];
mang size, num, tang;
void DFS(int u, int x) {
  size[u] = 1;
  tang[u] = x;
  num[u] = ++dem;
  for (int i = 0; i < a[u].size(); i++) {
    int v = a[u][i];
    if (tang[v] == 0) {
      DFS(v, x + 1);
      size[u] += size[v];
    }
  }
}
void nhap() {
  int u, v;
  cin >> n >> m;
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  DFS(1, 1);
}
void update(int node, int x, int type) {
  while (node < 200003) {
    c[type][node] += x;
    node += (node & (-node));
  }
}
void capnhat(int l, int r, int type, int x) {
  update(l, x, type);
  update(r + 1, -x, type);
}
long long getsum(int x, int type) {
  long long p = 0;
  while (x) {
    p += c[type][x];
    x -= (x & (-x));
  }
  return p;
}
void xuli() {
  int type, v, x, d;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d", &v);
      printf("%lld\n", 0ll + getsum(150005, 0) - getsum(tang[v] - 1, 0) +
                           getsum(num[v], 1));
    } else {
      scanf("%d%d%d", &v, &x, &d);
      if (v == 1) {
        update(d + 1, x, 0);
        continue;
      }
      if (tang[v] - 1 > d) {
        int l = num[v] - d;
        int r = min(num[v] + size[v] - 1, num[v] + d);
        capnhat(l, r, 1, x);
        continue;
      }
      int l = d - tang[v] + 1;
      update(l + 1, x, 0);
      int dau = num[v] - tang[v] + 1 + l + 1;
      int cuoi = min(num[v] + size[v] - 1, num[v] + d);
      if (dau <= cuoi) capnhat(dau, cuoi, 1, x);
    }
  }
}
int main() {
  nhap();
  xuli();
}
