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 N = 1e5 + 10;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 10;
int mark[N], H[N], p[N], sz[N], tot[4 * N], lz[4 * N];
vector<int> adj[N], seg[N], lzy[N];
void dfs(int v, int x) {
  mark[v] = 1;
  p[v] = x;
  sz[x]++;
  for (int u : adj[v]) {
    if (mark[u]) continue;
    H[u] = H[v] + 1;
    dfs(u, x);
  }
}
void shift(int c, int id, int l, int r) {
  seg[c][id] += lzy[c][id];
  if (r - l < 2) {
    lzy[c][id] = 0;
    return;
  }
  lzy[c][2 * id] += lzy[c][id];
  lzy[c][2 * id + 1] += lzy[c][id];
  lzy[c][id] = 0;
}
void add(int c, int val, int l, int r, int s, int e, int id) {
  shift(c, id, s, e);
  if (e <= l || r <= s) return;
  if (l <= s && e <= r) {
    lzy[c][id] += val;
    return;
  }
  int m = (s + e) / 2;
  add(c, val, l, r, s, m, 2 * id);
  add(c, val, l, r, m, e, 2 * id + 1);
  seg[c][id] = seg[c][2 * id] + seg[c][2 * id + 1];
}
int get(int c, int x, int s, int e, int id) {
  shift(c, id, s, e);
  if (e - s == 1) return seg[c][id];
  int m = (s + e) / 2;
  if (x < m) return get(c, x, s, m, 2 * id);
  return get(c, x, m, e, 2 * id + 1);
}
void shift1(int id, int l, int r) {
  tot[id] += lz[id];
  if (r - l < 2) {
    lz[id] = 0;
    return;
  }
  lz[2 * id] += lz[id];
  lz[2 * id + 1] += lz[id];
  lz[id] = 0;
}
void add1(int x, int l, int r, int s, int e, int id) {
  shift1(id, s, e);
  if (e <= l || r <= s) return;
  if (l <= s && e <= r) {
    lz[id] += x;
    return;
  }
  int m = (s + e) / 2;
  add1(x, l, r, s, m, 2 * id);
  add1(x, l, r, m, e, 2 * id + 1);
  tot[id] = tot[2 * id] + tot[2 * id + 1];
}
int get1(int x, int l, int r, int id) {
  shift1(id, l, r);
  if (r - l == 1) return tot[id];
  int m = (l + r) / 2;
  if (x < m) return get1(x, l, m, 2 * id);
  return get1(x, m, r, 2 * id + 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int t = 1;
  mark[1] = true;
  for (int v : adj[1]) dfs(v, t++);
  for (int i = 1; i < t; i++)
    seg[i].resize(4 * sz[i] + 10), lzy[i].resize(4 * sz[i] + 10);
  int root = 0;
  while (q--) {
    cin >> t;
    if (t == 0) {
      int v, x, d;
      cin >> v >> x >> d;
      if (v == 1) {
        root += x;
        add1(x, 0, d, 0, N, 1);
      } else {
        if (H[v] + 1 <= d) root += x;
        if (H[v] + 1 < d) {
          add1(x, 0, d - H[v] - 1, 0, N, 1);
          add(p[v], x, d - H[v] - 1, min(sz[p[v]], H[v] + d + 1), 0, sz[p[v]],
              1);
        } else {
          add(p[v], x, max(0, H[v] - d), min(sz[p[v]], H[v] + d + 1), 0,
              sz[p[v]], 1);
        }
      }
    } else {
      int v;
      cin >> v;
      if (v == 1)
        cout << root << '\n';
      else
        cout << get(p[v], H[v], 0, sz[p[v]], 1) + get1(H[v], 0, N, 1) << '\n';
    }
  }
  return 0;
}
