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 long long maxn = 1e5 + 10;
long long sz[maxn], pos[maxn], chain[maxn];
vector<long long> adj[maxn];
struct Tree {
  vector<long long> a;
  void init(long long n) {
    a.resize(n);
    fill(a.begin(), a.end(), 0);
  }
  void add(long long x, long long v) {
    x++;
    for (; x < a.size(); x += x & -x) a[x] += v;
  }
  long long query(long long x) {
    x++;
    long long ret = 0;
    for (; x; x -= x & -x) ret += a[x];
    return ret;
  }
} bit[maxn], G;
void dfs(long long u, long long par, long long c) {
  chain[u] = c;
  pos[u] = pos[par] + 1;
  sz[c]++;
  for (auto v : adj[u]) {
    if (v == par) continue;
    dfs(v, u, c);
  }
}
int main() {
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  long long c = 0;
  for (auto v : adj[1]) {
    dfs(v, 1, ++c);
  }
  for (long long i = 1; i <= c; i++) bit[i].init(sz[i] + 5);
  G.init(maxn);
  while (q--) {
    long long k;
    cin >> k;
    if (k == 0) {
      long long v, x, d;
      cin >> v >> x >> d;
      if (pos[v] == 0) {
        G.add(0, x);
        G.add(d + 1, -x);
        continue;
      }
      if (d >= pos[v]) {
        long long left = d - pos[v];
        long long l = left + 1;
        long long r = min(sz[chain[v]], pos[v] + d);
        if (l <= sz[chain[v]] && l <= r) {
          bit[chain[v]].add(l, x);
          bit[chain[v]].add(r + 1, -x);
        }
        G.add(0, x);
        G.add(left + 1, -x);
      } else {
        long long l = pos[v] - d;
        long long r = min(sz[chain[v]], pos[v] + d);
        bit[chain[v]].add(l, x);
        bit[chain[v]].add(r + 1, -x);
      }
    } else {
      long long v;
      cin >> v;
      if (v == 1)
        cout << G.query(0) << endl;
      else {
        long long ans = G.query(pos[v]) + bit[chain[v]].query(pos[v]);
        cout << ans << endl;
      }
    }
  }
  return 0;
}
