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 = 100000 + 10;
struct BIT {
  int n, *a;
  BIT(int m) {
    n = m + 2;
    a = new int[n];
    fill(a, a + n, 0);
  }
  void add(int p, int x) {
    while (p < n) {
      a[p] += x;
      p += p & -p;
    }
  }
  int sum(int p) {
    int res = 0;
    while (p > 0) {
      res += a[p];
      p -= p & -p;
    }
    return res;
  }
};
BIT* bit[MAXN];
int id[MAXN], dis[MAXN];
vector<int> e[MAXN];
int main() {
  int n, m, a, b, v;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    --a, --b;
    e[a].push_back(b), e[b].push_back(a);
  }
  fill(dis, dis + n, -1);
  int root = 0;
  n = e[0].size();
  id[0] = n;
  dis[0] = 0;
  for (int i = 0; i < n; i++) {
    int k = 1;
    for (int j = e[0][i];; j = dis[e[j][0]] == -1 ? e[j][0] : e[j][1]) {
      id[j] = i;
      dis[j] = k++;
      if (e[j].size() == 1) {
        break;
      }
    }
    bit[i] = new BIT(k);
  }
  bit[n] = new BIT(MAXN);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &a, &v);
    v--;
    if (a == 0) {
      scanf("%d%d", &a, &b);
      if (!v) {
        root += a;
        bit[n]->add(1, a);
        bit[n]->add(b + 1, -a);
      } else if (b < dis[v]) {
        bit[id[v]]->add(dis[v] - b, a);
        bit[id[v]]->add(dis[v] + b + 1, -a);
      } else {
        int left = b - dis[v];
        root += a;
        bit[n]->add(1, a);
        bit[n]->add(left + 1, -a);
        bit[id[v]]->add(left + 1, a);
        bit[id[v]]->add(dis[v] + b + 1, -a);
      }
    } else {
      if (v == 0)
        b = root;
      else
        b = bit[id[v]]->sum(dis[v]) + bit[n]->sum(dis[v]);
      printf("%d\n", b);
    }
  }
  return 0;
}
