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;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline string get() {
  char ch[1000010];
  scanf("%s", ch);
  return ch;
}
const int MAX_LG = 21;
const long long maxn = 1e5 + 10;
const long long base = 29;
const long long mod = 1e9 + 7;
const long long INF = 1e9 + 100;
vector<long long> g[maxn];
long long h[maxn];
vector<long long> fen;
vector<long long> fen2[maxn];
long long sz[maxn];
long long id[maxn];
inline void dfs(long long v, long long pr = -1) {
  sz[v] = 1;
  if (v != 1) id[v] = (pr == 1 ? v : id[pr]);
  for (auto u : g[v]) {
    if (u == pr) continue;
    h[u] = h[v] + 1;
    dfs(u, v);
    sz[v] += sz[u];
  }
}
inline void add(vector<long long> *f, long long value, long long p) {
  long long sz = (*f).size();
  for (; p < sz; p += p & -p) (*f)[p] += value;
}
inline long long get(vector<long long> *f, long long p) {
  if (p <= 0) return 0;
  long long ret = 0;
  for (; p; p -= p & -p) ret += (*f)[p];
  return ret;
}
long long res;
int32_t main() {
  fen.resize(maxn);
  long long n = in(), q = in();
  for (long long i = 0; i < n - 1; i++) {
    long long v = in(), u = in();
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(1);
  for (auto u : g[1]) fen2[u].resize(sz[u] + 5);
  for (long long i = 0; i < q; i++) {
    long long tp = in();
    if (tp == 0) {
      long long node = in(), value = in(), di = in();
      if (di >= h[node]) res += value;
      if (di > h[node]) {
        long long di2 = di - h[node];
        add(&fen, value, di2);
        if (node == 1) continue;
        long long di3 = min(h[node] + di, sz[id[node]]);
        if (di3 > di2) {
          add(&fen2[id[node]], value, di3);
          add(&fen2[id[node]], -value, di2);
        }
      } else {
        long long di2 = min(h[node] + di, sz[id[node]]);
        long long di3 = h[node] - di - 1;
        add(&fen2[id[node]], value, di2);
        if (di3 > 0) add(&fen2[id[node]], -value, di3);
      }
    } else {
      long long node = in();
      if (node == 1) {
        cout << res << "\n";
        continue;
      }
      cout << get(&fen, maxn - 1) - get(&fen, h[node] - 1) +
                  get(&fen2[id[node]], (long long)fen2[id[node]].size() - 1) -
                  get(&fen2[id[node]], h[node] - 1)
           << "\n";
    }
  }
}
