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 mark[100010], level[100010], pos[100010], length[100010], id, deep;
vector<int> e[100010];
class BIT {
 private:
  int *sum, n;

 public:
  void init(int _n) {
    n = _n;
    sum = new int[n + 1];
    memset(sum, 0, (n + 1) * sizeof(int));
  }
  int query(int i) {
    int ans = 0;
    for (; i <= n; i += i & -i) ans += sum[i];
    return ans;
  }
  void add(int i, int x) {
    for (; i > 0; i -= i & -i) sum[i] += x;
  }
  void update(int l, int r, int x) {
    add(r, x);
    add(l - 1, -x);
  }
} tree, *chain;
int dfs(int step, int u, int pre) {
  int i, n = e[u].size(), v;
  level[u] = step + 1;
  mark[u] = id;
  pos[u] = step;
  for (i = 0; i < n; i++) {
    v = e[u][i];
    if (v == pre) continue;
    return dfs(step + 1, v, u);
  }
  return step;
}
void pretreat() {
  int i, n = e[1].size(), len, v;
  chain = new BIT[n];
  level[1] = 1;
  for (i = 0; i < n; i++) {
    id = i;
    v = e[1][i];
    len = dfs(1, v, 1);
    length[i] = len;
    deep = max(len, deep);
    chain[i].init(len);
  }
  tree.init(++deep);
}
int query(int v) {
  int ans = 0;
  ans = tree.query(level[v]);
  if (v != 1) {
    ans += chain[mark[v]].query(pos[v]);
  }
  return ans;
}
void update(int v, int x, int d) {
  int l, r;
  if (v == 1) {
    if (deep >= 1 + d)
      r = 1 + d;
    else
      r = deep;
    tree.update(1, r, x);
    return;
  }
  if (length[mark[v]] >= pos[v] + d)
    r = pos[v] + d;
  else
    r = length[mark[v]];
  if (1 <= pos[v] - d)
    l = pos[v] - d;
  else
    l = 1;
  chain[mark[v]].update(l, r, x);
  d -= level[v] - 1;
  if (d >= 0) {
    if (deep >= 1 + d)
      r = 1 + d;
    else
      r = deep;
    tree.update(1, r, x);
    if (r >= 2) {
      if (length[mark[v]] >= r - 1)
        r = r - 1;
      else
        r = length[mark[v]];
      chain[mark[v]].update(1, r, -x);
    }
  }
}
int main() {
  int i, n, q, x, y;
  scanf("%d %d", &n, &q);
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d", &x, &y);
    e[x].push_back(y);
    e[y].push_back(x);
  }
  pretreat();
  int op, v, d;
  while (q--) {
    scanf("%d %d", &op, &v);
    if (op) {
      printf("%d\n", query(v));
    } else {
      scanf("%d %d", &x, &d);
      update(v, x, d);
    }
  }
  return 0;
}
