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 MAX = 200001;
vector<int> adj[MAX];
int n_tree[MAX], h_tree[MAX], pos[MAX];
vector<vector<long long> > tree;
int N;
bool mark[MAX];
long long one = 0;
void build() {
  int i, u, v;
  queue<int> Q;
  mark[1] = true;
  tree.resize(adj[1].size() + 1);
  Q.push(1);
  while (!Q.empty()) {
    u = Q.front(), Q.pop();
    for (i = 0; i < (int)adj[u].size(); i++) {
      v = adj[u][i];
      if (!mark[v]) {
        Q.push(v);
        mark[v] = true;
        pos[v] = pos[u] + 1;
        n_tree[v] = (u == 1) ? i + 1 : n_tree[u];
        h_tree[n_tree[v]]++;
      }
    }
  }
  for (i = 0; i < (int)adj[1].size(); i++)
    tree[i + 1].resize(h_tree[i + 1] + 1, 0);
  h_tree[0] = N;
  tree[0].resize(N + 1, 0);
}
void update(int branch, int pos, long long val) {
  for (pos = max(pos, 1); pos <= h_tree[branch]; pos += pos & -pos)
    tree[branch][pos] += val;
}
long long read(int branch, int pos) {
  long long ret = 0;
  for (; pos > 0; pos -= pos & -pos) ret += tree[branch][pos];
  return ret;
}
long long query(int u) {
  if (u == 1) return one;
  return read(n_tree[u], pos[u]) + read(0, pos[u]);
}
void update_root(int val, int d) {
  if (d <= 0) return;
  update(0, 1, val);
  update(0, 1 + d, -val);
}
void actualiza(int u, long long val, int d) {
  if (1 == u) {
    one += val;
    update_root(val, d);
    return;
  }
  if (pos[u] - d <= 0) one += val;
  update(n_tree[u], pos[u] - d, val);
  update(n_tree[u], pos[u] + d + 1, -val);
  int comun = max(0, d - pos[u]);
  if (comun) {
    update_root(val, comun);
    update(n_tree[u], 1, -val);
    update(n_tree[u], comun + 1, val);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int i, M, u, v, type, d, val;
  cin >> N >> M;
  for (i = 0; i < N - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  build();
  while (M--) {
    cin >> type;
    if (type) {
      cin >> u;
      cout << query(u) << endl;
    } else {
      cin >> u >> val >> d;
      actualiza(u, val, d);
    }
  }
  return 0;
}
