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 maxn = 100 * 1000 + 10;
int n, q, u, v, x, d, s[maxn], f[maxn], sum[4 * maxn], dis[maxn],
    data[4 * maxn], t, jvb[4 * maxn];
vector<int> vec[maxn];
inline void dfs(int x, int p) {
  s[x] = t;
  t++;
  if (p != -1) dis[x] = dis[p] + 1;
  for (int i = 0; i < ((int(vec[x].size()))); i++) {
    if (vec[x][i] != p) dfs(vec[x][i], x);
  }
  f[x] = t;
}
inline void push(int x, int l, int r) {
  int mid = (l + r) / 2;
  sum[2 * x + 1] += (mid - l) * data[x];
  sum[2 * x + 2] += (r - mid) * data[x];
  data[2 * x + 1] += data[x];
  data[2 * x + 2] += data[x];
  data[x] = 0;
}
inline void upd(int x, int l, int r, int ql, int qr, int val) {
  if (l >= qr || ql >= r) return;
  if (ql <= l && r <= qr) {
    data[x] += val;
    sum[x] += (r - l) * val;
    return;
  }
  int mid = (l + r) / 2;
  push(x, l, r);
  upd(2 * x + 1, l, mid, ql, qr, val);
  upd(2 * x + 2, mid, r, ql, qr, val);
  sum[x] = sum[2 * x + 1] + sum[2 * x + 2];
}
inline void add(int x, int l, int r, int a, int val) {
  if (r - l == 1) {
    jvb[x] += val;
    return;
  }
  int mid = (l + r) / 2;
  if (a < mid)
    add(2 * x + 1, l, mid, a, val);
  else
    add(2 * x + 2, mid, r, a, val);
  jvb[x] = jvb[2 * x + 1] + jvb[2 * x + 2];
}
inline int req(int x, int l, int r, int a) {
  if (r - l == 1) return sum[x];
  int mid = (l + r) / 2;
  push(x, l, r);
  if (a < mid)
    return req(2 * x + 1, l, mid, a);
  else
    return req(2 * x + 2, mid, r, a);
}
inline int ans(int x, int l, int r, int ql, int qr) {
  if (l >= qr || ql >= r) return 0;
  if (ql <= l && r <= qr) return jvb[x];
  int mid = (l + r) / 2;
  return ans(2 * x + 1, l, mid, ql, qr) + ans(2 * x + 2, mid, r, ql, qr);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--;
    v--;
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  dfs(0, -1);
  for (int i = 0; i < q; i++) {
    cin >> t >> v;
    v--;
    if (t == 0) {
      cin >> x >> d;
      if (dis[v] > d)
        upd(0, 0, n, s[v] - d, min(f[v], s[v] + d + 1), x);
      else {
        add(0, 0, n, d - dis[v], x);
        upd(0, 0, n, min(s[v] - 2 * dis[v] + d + 1, f[v]),
            min(f[v], s[v] + d + 1), x);
      }
    } else
      cout << req(0, 0, n, s[v]) + ans(0, 0, n, dis[v], n) << endl;
  }
  return 0;
}
