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;
template <class T>
void rd(T &x) {
  x = 0;
  int f = 1;
  char c;
  while (!isdigit(c = getchar()))
    if (c == '-') f = -1;
  do {
    x = (x << 3) + (x << 1) + (c ^ 48);
  } while (isdigit(c = getchar()));
  x *= f;
}
const int N = 1e5 + 5;
int n, m, cnt, tot, ans, dp[N], dep[N], num[N], l[N], r[N], val[N << 2][2];
vector<int> G[N];
void dfs(int x, int fa) {
  num[x] = ++cnt;
  dep[x] = dep[fa] + 1;
  for (auto it : G[x])
    if (it != fa) dfs(it, x);
}
void modify(int p, int l, int r, int L, int R, int v, int dis) {
  if (l > R || r < L) return;
  if (L <= l && R >= r) {
    val[p][dis] += v;
    return;
  }
  int mid = l + r >> 1;
  modify(p << 1, l, mid, L, R, v, dis);
  modify(p << 1 | 1, mid + 1, r, L, R, v, dis);
}
void query(int p, int l, int r, int v, int dis) {
  ans += val[p][dis];
  if (l == r) return;
  int mid = l + r >> 1;
  if (v <= mid)
    query(p << 1, l, mid, v, dis);
  else
    query(p << 1 | 1, mid + 1, r, v, dis);
}
int main() {
  rd(n), rd(m);
  for (int i = 1, x, y; i < n; i++)
    rd(x), rd(y), G[x].push_back(y), G[y].push_back(x);
  num[1] = ++cnt, dep[1] = 1;
  for (auto it : G[1]) {
    int tp = cnt + 1;
    dfs(it, 1);
    for (int i = tp; i <= cnt; i++) l[i] = tp, r[i] = cnt;
  }
  for (int i = 1; i <= m; i++) {
    int opt, v, x, d;
    rd(opt), rd(v);
    if (!opt) {
      rd(x), rd(d);
      if (v != 1)
        modify(1, 1, n, max(l[num[v]], num[v] - d), min(r[num[v]], num[v] + d),
               x, 0);
      if (dep[v] <= d + 1) {
        modify(1, 1, n, 1, d - dep[v] + 2, x, 1);
        if (dep[v] <= d)
          modify(1, 1, n, l[num[v]], min(r[num[v]], l[num[v]] + d - dep[v]), -x,
                 0);
      }
    } else {
      ans = 0;
      query(1, 1, n, num[v], 0);
      query(1, 1, n, dep[v], 1);
      printf("%d\n", ans);
    }
  }
  return 0;
}
