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 + 100;
const long long int mod = 998244353;
const int inf = 1e9 + 100;
int st[N], com[N], par[N], l[N], r[N], h[N], ti = 1;
long long int seg[2][4 * N], lazy[2][4 * N];
vector<int> adj[N];
void dfs(int u, int i) {
  st[u] = ti++;
  com[u] = i;
  for (auto x : adj[u]) {
    if (x != par[u]) {
      h[x] = h[u] + 1;
      par[x] = u;
      dfs(x, i);
    }
  }
  if (adj[u].size() == 1) r[i] = u;
}
inline void upd(int ind) {
  seg[0][ind * 2] += lazy[0][ind];
  seg[0][ind * 2 + 1] += lazy[0][ind];
  lazy[0][ind * 2] += lazy[0][ind];
  lazy[0][ind * 2 + 1] += lazy[0][ind];
  lazy[0][ind] = 0;
  seg[1][ind * 2] += lazy[1][ind];
  seg[1][ind * 2 + 1] += lazy[1][ind];
  lazy[1][ind * 2] += lazy[1][ind];
  lazy[1][ind * 2 + 1] += lazy[1][ind];
  lazy[1][ind] = 0;
}
void add(int x, int L, int R, int val, int b = 0, int e = N, int ind = 1) {
  if (R <= b || e <= L) return;
  if (L <= b && e <= R) {
    seg[x][ind] += val;
    lazy[x][ind] += val;
    return;
  }
  upd(ind);
  int mid = (b + e) / 2;
  add(x, L, R, val, b, mid, ind * 2);
  add(x, L, R, val, mid, e, ind * 2 + 1);
  seg[x][ind] = seg[x][ind * 2] + seg[x][ind * 2 + 1];
}
long long int get(int x, int L, int R, int b = 0, int e = N, int ind = 1) {
  if (R <= b || e <= L) return 0;
  if (L <= b && e <= R) {
    return seg[x][ind];
  }
  upd(ind);
  int mid = (b + e) / 2;
  return get(x, L, R, b, mid, ind * 2) + get(x, L, R, mid, e, ind * 2 + 1);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 0; i <= n; i++) com[i] = -1;
  for (int i = 0; i < adj[0].size(); i++) {
    int x = adj[0][i];
    l[i] = x;
    h[x] = 1;
    par[x] = 0;
    dfs(x, i);
    l[i] = st[l[i]];
    r[i] = st[r[i]];
  }
  while (q--) {
    int t;
    cin >> t;
    if (t == 0) {
      int u, d, val;
      cin >> u >> val >> d;
      u--;
      if (u == 0) {
        add(1, d, d + 1, val);
        continue;
      }
      int c = com[u];
      add(0, st[u], min(r[c], st[u] + d) + 1, val);
      add(0, max(l[c], st[u] - d), st[u], val);
      int k = d - h[u];
      if (k < 0) continue;
      add(1, k, k + 1, val);
      add(0, l[c], min(r[c] + 1, l[c] + k), -val);
    } else {
      int u;
      cin >> u;
      u--;
      cout << get(0, st[u], st[u] + 1) + get(1, h[u], n + 1) << endl;
    }
  }
  return 0;
}
