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>
const int MN = 1e5 + 4;
int N, Q, ecnt, ch[MN], depth[MN], len[MN];
long long rt, bit1[MN], *bit2[MN];
std::vector<int> adj[MN];
inline void upd(long long *bit, int ind, int cap, long long val) {
  for (; ind < cap; ind += ind & -ind) bit[ind] += val;
}
inline long long qry(long long *bit, int ind) {
  long long res = 0;
  for (; ind; ind -= ind & -ind) res += bit[ind];
  return res;
}
void dfs1(int src, int p, int d, int b) {
  ++ecnt;
  ch[src] = b;
  depth[src] = d++;
  for (int i : adj[src]) {
    if (i == p) continue;
    dfs1(i, src, d, b);
  }
}
int main() {
  scanf("%d%d", &N, &Q);
  for (int i = 1, a, b; i < N; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].emplace_back(b);
    adj[b].emplace_back(a);
  }
  int cnt = 0;
  for (int i : adj[1]) {
    dfs1(i, 1, 1, ++cnt);
    len[cnt] = ecnt + 1;
    bit2[cnt] = (long long *)calloc(ecnt + 1, sizeof(long long));
    ecnt = 0;
  }
  for (int i = 0, a, b, c, d; i < Q; ++i) {
    scanf("%d", &a);
    if (a) {
      scanf("%d", &b);
      if (b == 1)
        printf("%lld\n", rt);
      else
        printf("%lld\n", qry(bit1, depth[b]) + qry(bit2[ch[b]], depth[b]));
    } else {
      scanf("%d%d%d", &b, &c, &d);
      if (b == 1) {
        upd(bit1, 1, MN, c);
        upd(bit1, d + 1, MN, -c);
        rt += c;
      } else {
        if (d < depth[b]) {
          upd(bit2[ch[b]], depth[b] - d, len[ch[b]], c);
          upd(bit2[ch[b]], depth[b] + d + 1, len[ch[b]], -c);
        } else {
          upd(bit2[ch[b]], 1, len[ch[b]], c);
          upd(bit2[ch[b]], depth[b] + d + 1, len[ch[b]], -c);
          if (depth[b] != d) {
            upd(bit1, 1, MN, c);
            upd(bit1, d - depth[b] + 1, MN, -c);
            upd(bit2[ch[b]], 1, len[ch[b]], -c);
            upd(bit2[ch[b]], d - depth[b] + 1, len[ch[b]], c);
          }
          rt += c;
        }
      }
    }
  }
}
