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;
int n, q, x, y, z, zz, belong[200005], deep[200005], size[200005], root[200005],
    treesize, ANS;
vector<int> e[200005];
struct segtree {
  int l, r, s, mark;
  int mid() { return (l + r) >> 1; }
} tree[16 * 200005];
int dfs(int x, int f, int id, int de) {
  belong[x] = id, deep[x] = de, size[x] = 1;
  for (int i = 0, l = e[x].size(); i < l; i++)
    if (e[x][i] != f) {
      if (x == 1)
        size[x] += dfs(e[x][i], x, i + 1, de + 1);
      else
        size[x] += dfs(e[x][i], x, id, de + 1);
    }
  return size[x];
}
void buildtree(int l, int r, int xx, int delta) {
  int x = xx + delta;
  tree[x].l = l, tree[x].r = r;
  if (l == r) return;
  int mid = tree[x].mid();
  buildtree(l, mid, (xx << 1), delta);
  buildtree(mid + 1, r, ((xx << 1) + 1), delta);
}
void add_mark(int l, int r, int ad, int xx, int delta) {
  int x = xx + delta;
  if (tree[x].l >= l && tree[x].r <= r) {
    tree[x].mark += ad;
    return;
  }
  if (tree[x].mark) {
    tree[delta + (xx << 1)].mark += tree[x].mark,
        tree[delta + ((xx << 1) + 1)].mark += tree[x].mark;
    tree[x].mark = 0;
  }
  int mid = tree[x].mid();
  if (l <= mid) add_mark(l, r, ad, (xx << 1), delta);
  if (r > mid) add_mark(l, r, ad, ((xx << 1) + 1), delta);
}
int query(int pnt, int xx, int delta) {
  int x = xx + delta;
  if (tree[x].l == tree[x].r) {
    return tree[x].mark;
  }
  if (tree[x].mark) {
    tree[delta + (xx << 1)].mark += tree[x].mark,
        tree[delta + ((xx << 1) + 1)].mark += tree[x].mark;
    tree[x].mark = 0;
  }
  int mid = tree[x].mid();
  if (pnt <= mid)
    return query(pnt, (xx << 1), delta);
  else
    return query(pnt, ((xx << 1) + 1), delta);
}
int main() {
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    e[x].push_back(y), e[y].push_back(x);
  }
  dfs(1, 0, 0, 0);
  for (int i = 1, l = e[1].size(); i <= l; i++) {
    root[e[1][i - 1]] = treesize;
    treesize += size[e[1][i - 1]] * 4;
    buildtree(1, size[e[1][i - 1]], 1, root[e[1][i - 1]]);
  }
  root[0] = treesize;
  buildtree(1, 100005, 1, root[0]);
  for (int i = 1; i <= q; i++) {
    scanf("%d", &x);
    if (x == 0) {
      scanf("%d%d%d", &y, &z, &zz);
      int b = e[1][belong[y] - 1];
      if (y != 1) {
        int k = min(size[b] - deep[y], zz), kk = min(deep[y] - 1, zz);
        add_mark(deep[y] - kk, deep[y] + k, z, 1, root[b]);
        if (deep[y] <= zz) ANS += z;
        int tt = zz - deep[y];
        if (tt > 0) {
          add_mark(1, tt, z, 1, root[0]);
          add_mark(1, tt, -z, 1, root[b]);
        }
      }
      if (y == 1) {
        ANS += z;
        add_mark(1, zz, z, 1, root[0]);
      }
    } else {
      scanf("%d", &y);
      int b = e[1][belong[y] - 1];
      if (y == 1)
        cout << ANS << endl;
      else {
        printf("%d\n", query(deep[y], 1, root[b]) + query(deep[y], 1, root[0]));
      }
    }
  }
  return 0;
}
