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;
struct BIT {
 public:
  vector<int> t;
  int n;

 private:
  void add(int x, int v) {
    for (; x < n; x += x & -x) {
      t[x] += v;
    }
  }

 public:
  BIT(int sz) {
    t = vector<int>(sz);
    n = sz;
  }
  void add(int l, int r, int v) {
    ++l;
    ++r;
    add(l, v);
    add(r, -v);
  }
  int query(int x) {
    ++x;
    int ans = 0;
    for (; x; x -= x & -x) {
      ans += t[x];
    }
    return ans;
  }
};
const int N = 100100;
vector<BIT> bits;
vector<int> adj[N];
int my_bit[N];
int my_ind[N];
int n, q;
int dfs(int u, int p, int bit_ind, int ind) {
  my_bit[u] = bit_ind;
  my_ind[u] = ind;
  int res = ind;
  for (auto& v : adj[u]) {
    if (v == p) continue;
    res = dfs(v, u, bit_ind, ind + 1);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  bits.emplace_back(1);
  int mx = 5;
  for (auto& v : adj[1]) {
    int amt = dfs(v, 1, bits.size(), 2);
    bits.emplace_back(amt + 5);
    mx = max(mx, amt + 5);
  }
  BIT all(mx);
  my_ind[1] = 1;
  for (int qq = 0; qq < q; ++qq) {
    int type;
    cin >> type;
    if (type == 0) {
      int u, x, d;
      cin >> u >> x >> d;
      if (u == 1) {
        all.add(1, 2 + d, x);
      } else {
        int extra = 0;
        int lo = my_ind[u] - d;
        int hi = my_ind[u] + d + 1;
        if (lo < 2) {
          extra = 2 - lo;
          lo = 2;
        }
        bits[my_bit[u]].add(lo, hi, x);
        if (extra) {
          all.add(1, extra + 1, x);
          bits[my_bit[u]].add(1, extra + 1, -x);
        }
      }
    } else if (type == 1) {
      int u;
      cin >> u;
      int res = all.query(my_ind[u]);
      if (u != 1) {
        assert(my_ind[u] > 1);
        res += bits[my_bit[u]].query(my_ind[u]);
      }
      cout << res << "\n";
    }
  }
  return 0;
}
