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 = 100005;
int dep[maxn], n, tt, bid[maxn], did[maxn];
int lowbit(int x) { return x & -x; }
struct bit {
  int board[maxn];
  void modify(int p, int v) {
    for (; p <= n; p += lowbit(p)) board[p] += v;
  }
  int query(int p) {
    int res = 0;
    for (; p; p -= lowbit(p)) res += board[p];
    return res;
  }
};
bit ta, tb;
struct node {
  int id, val;
  node(int x, int y) : id(x), val(y) {}
  bool operator<(const node& a) const { return val < a.val; }
};
vector<int> v[maxn];
vector<node> so;
int tod[maxn], bot[maxn];
int dfs(int x, int f, int d) {
  dep[x] = bot[x] = d;
  did[x] = ++tt;
  so.push_back(node(x, d));
  tod[d]++;
  for (int i = 0; i < v[x].size(); i++) {
    int toit = v[x][i];
    if (toit != f) bot[x] = dfs(toit, x, d + 1);
  }
  return bot[x];
}
void pre() {
  dfs(1, 0, 0);
  sort(so.begin(), so.end());
  for (int i = 0; i < n; i++) bid[so[i].id] = i + 1;
  for (int i = 1; i < n; i++) tod[i] += tod[i - 1];
}
int main() {
  int q, a, b, c, d;
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  pre();
  while (q--) {
    scanf("%d", &a);
    if (a == 1) {
      scanf("%d", &b);
      printf("%d\n", ta.query(did[b]) + tb.query(bid[b]));
    } else {
      scanf("%d%d%d", &b, &c, &d);
      if (dep[b] <= d) {
        tb.modify(1, c);
        tb.modify(tod[d - dep[b]] + 1, -c);
        int tp = dep[b] - (d - dep[b] + 1);
        int l = did[b] - tp, r = did[b] + min(d, bot[b] - dep[b]);
        if (l <= r) {
          ta.modify(l, c);
          ta.modify(r + 1, -c);
        }
      } else {
        int l = did[b] - d, r = did[b] + min(d, bot[b] - dep[b]);
        ta.modify(l, c);
        ta.modify(r + 1, -c);
      }
    }
  }
}
