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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int Maxn = 100005;
int n, q;
vector<int> neigh[Maxn];
vector<long long> RA;
vector<int> cnt, tsiz;
vector<vector<long long> > BIT, stree;
int r[Maxn], no[Maxn];
void Create(int p, int v, int R) {
  r[v] = R;
  no[v] = ++cnt[R];
  for (int i = 0; i < neigh[v].size(); i++) {
    int u = neigh[v][i];
    if (u != p) Create(v, u, R);
  }
}
void addBIT(vector<long long> &B, int x, int val) {
  for (int i = x; i > 0; i -= i & -i) B[i] += val;
}
void addST(vector<long long> &st, int v, int l, int r, int a, int b, int x) {
  if (l == a && r == b)
    st[v] += x;
  else {
    int mid = l + r >> 1;
    if (a <= mid) addST(st, 2 * v, l, mid, a, min(mid, b), x);
    if (mid < b) addST(st, 2 * v + 1, mid + 1, r, max(mid + 1, a), b, x);
  }
}
long long getBIT(const vector<long long> &B, int x) {
  long long res = 0ll;
  for (int i = x; i < B.size(); i += i & -i) res += B[i];
  return res;
}
long long getST(const vector<long long> &st, int v, int l, int r, int a) {
  if (l == a && r == a)
    return st[v];
  else {
    long long res = st[v];
    int mid = l + r >> 1;
    if (a <= mid)
      res += getST(st, 2 * v, l, mid, a);
    else
      res += getST(st, 2 * v + 1, mid + 1, r, a);
    return res;
  }
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    neigh[u].push_back(v);
    neigh[v].push_back(u);
  }
  cnt.assign(neigh[1].size(), 0);
  tsiz.resize(neigh[1].size());
  BIT.resize(neigh[1].size());
  stree.resize(neigh[1].size());
  for (int i = 0; i < neigh[1].size(); i++) {
    Create(1, neigh[1][i], i);
    BIT[i].assign(cnt[i] + 5, 0);
    tsiz[i] =
        int(pow(2.0, floor(log(double(cnt[i] + 5)) / log(2.0)) + 2.0) + 5);
    stree[i].assign(tsiz[i], 0);
  }
  RA.assign(Maxn, 0);
  while (q--) {
    int com, v;
    scanf("%d %d", &com, &v);
    if (com == 0) {
      int x, d;
      scanf("%d %d", &x, &d);
      if (v == 1)
        addBIT(RA, d + 1, x);
      else {
        addST(stree[r[v]], 1, 1, cnt[r[v]], max(1, no[v] - d),
              min(cnt[r[v]], no[v] + d), x);
        if (d >= no[v]) {
          addBIT(RA, d - no[v] + 1, x);
          addBIT(BIT[r[v]], min(cnt[r[v]], d - no[v]) + 1, -x);
        }
      }
    } else if (v == 1)
      printf("%I64d\n", getBIT(RA, 1));
    else
      printf("%I64d\n", getBIT(RA, no[v] + 1) + getBIT(BIT[r[v]], no[v] + 1) +
                            getST(stree[r[v]], 1, 1, cnt[r[v]], no[v]));
  }
  return 0;
}
