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 M = 100005;
int read() {
  int x = 0, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9') {
    if (c == '-') f = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
int n, m, cnt, tot, ans, f[M], dep[M], num[M], l[M], r[M];
int val[4 * M][2];
struct edge {
  int v, next;
} e[2 * M];
void dfs(int u, int fa) {
  num[u] = ++cnt;
  dep[u] = dep[fa] + 1;
  for (int i = f[u]; i; i = e[i].next)
    if (e[i].v ^ fa) dfs(e[i].v, u);
}
void modify(int i, int l, int r, int L, int R, int v, int d) {
  if (l > R || L > r) return;
  if (L <= l && r <= R) {
    val[i][d] += v;
    return;
  }
  int mid = (l + r) >> 1;
  modify(i << 1, l, mid, L, R, v, d);
  modify(i << 1 | 1, mid + 1, r, L, R, v, d);
}
void ask(int i, int l, int r, int id, int d) {
  ans += val[i][d];
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (id <= mid)
    ask(i << 1, l, mid, id, d);
  else
    ask(i << 1 | 1, mid + 1, r, id, d);
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    e[++tot] = edge{v, f[u]}, f[u] = tot;
    e[++tot] = edge{u, f[v]}, f[v] = tot;
  }
  num[1] = ++cnt;
  dep[1] = 1;
  for (int i = f[1]; i; i = e[i].next) {
    int st = cnt + 1;
    dfs(e[i].v, 1);
    for (int j = st; j <= cnt; j++) l[j] = st, r[j] = cnt;
  }
  for (int i = 1; i <= m; i++) {
    int op = read(), v = read();
    if (op == 0) {
      int x = read(), d = read();
      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;
      ask(1, 1, n, num[v], 0);
      ask(1, 1, n, dep[v], 1);
      printf("%d\n", ans);
    }
  }
}
