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 N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int x, d, w, v, n, q, i, u;
int belong[N], S[N], ch[N], whch[N];
vector<int> first[N], V[N * 4], lazy[N * 4];
void f(int x, int ata, int s, int b) {
  S[x] = s;
  ch[x] = 1;
  belong[x] = b;
  int i;
  for (i = 0; i < V[x].size(); i++)
    if (V[x][i] != ata) {
      f(V[x][i], x, s + 1, b);
      ch[x] += ch[V[x][i]];
    }
}
void push(int root, int bas, int son, int b) {
  lazy[b][(root + root)] += lazy[b][root];
  lazy[b][(root + root + 1)] += lazy[b][root];
  first[b][(root + root)] += lazy[b][root] * (((bas + son) / 2) - bas + 1);
  first[b][(root + root + 1)] += lazy[b][root] * (son - ((bas + son) / 2));
  lazy[b][root] = 0;
}
int update(int root, int bas, int son, int x, int y, int v, int b) {
  if (bas > y or son < x) return first[b][root];
  if (x <= bas and son <= y) {
    first[b][root] += v * (son - bas + 1);
    lazy[b][root] += v;
    return first[b][root];
  }
  if (lazy[b][root]) push(root, bas, son, b);
  return first[b][root] =
             update((root + root), bas, ((bas + son) / 2), x, y, v, b) +
             update((root + root + 1), ((bas + son) / 2) + 1, son, x, y, v, b);
}
int query(int root, int bas, int son, int x, int b) {
  if (bas > x or son < x) return 0;
  if (bas == son) return first[b][root];
  if (lazy[b][root]) push(root, bas, son, b);
  return query((root + root), bas, ((bas + son) / 2), x, b) +
         query((root + root + 1), ((bas + son) / 2) + 1, son, x, b);
}
int main() {
  scanf("%d %d", &n, &q);
  for (i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    V[u].push_back(v);
    V[v].push_back(u);
  }
  for (i = 0; i < V[1].size(); i++) {
    f(V[1][i], 1, 1, V[1][i]);
    whch[V[1][i]] = i;
  }
  for (i = 0; i < V[1].size(); i++) {
    first[i].resize(ch[V[1][i]] * 4);
    lazy[i].resize(ch[V[1][i]] * 4);
  }
  first[V[1].size()].resize(N * 4);
  lazy[V[1].size()].resize(N * 4);
  for (i = 1; i <= q; i++) {
    scanf("%d %d", &w, &v);
    if (w == 1) {
      int ans;
      if (v != 1)
        ans = query(1, 1, ch[belong[v]], S[v], whch[belong[v]]) +
              query(1, 1, N, S[v] + 1, V[1].size());
      else
        ans = query(1, 1, N, 1, V[1].size());
      printf("%d\n", ans);
    } else {
      scanf("%d %d", &x, &d);
      if (v == 1) {
        update(1, 1, N, 1, d + 1, x, V[1].size());
        continue;
      }
      update(1, 1, ch[belong[v]], max(1, S[v] - d),
             min(ch[belong[v]], S[v] + d), x, whch[belong[v]]);
      if (d - S[v] + 1 > 0) {
        update(1, 1, N, 1, min(N, d - S[v] + 1), x, V[1].size());
        if (d - S[v])
          update(1, 1, ch[belong[v]], 1, min(ch[belong[v]], d - S[v]), -x,
                 whch[belong[v]]);
      }
    }
  }
}
