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
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF276E extends PrintWriter {
	CF276E() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF276E o = new CF276E(); o.main(); o.flush();
	}

	int[] next, jj; int $ = 1;
	int[] ao, hh, dd;
	int link(int l, int j) {
		next[$] = l; jj[$] = j;
		return $++;
	}
	void init(int n, int m) {
		next = new int[1 + m * 2]; jj = new int[1 + m * 2];
		ao = new int[n]; hh = new int[n]; dd = new int[n];
	}
	int d_;
	void dfs(int p, int i, int h, int d) {
		hh[i] = h; dd[i] = d;
		d_ = Math.max(d_, d);
		for (int l = ao[i]; l != 0; l = next[l]) {
			int j = jj[l];
			if (j != p)
				dfs(i, j, h, d + 1);
		}
	}
	void update(int[] aa, int i, int x) {
		while (i < aa.length) {
			aa[i] += x;
			i |= i + 1;
		}
	}
	int query(int[] aa, int i) {
		int x = 0;
		while (i >= 0) {
			x += aa[i];
			i &= i + 1; i--;
		}
		return x;
	}
	void main() {
		int n = sc.nextInt();
		int q = sc.nextInt();
		init(n, n - 1);
		for (int h = 0; h < n - 1; h++) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			ao[i] = link(ao[i], j);
			ao[j] = link(ao[j], i);
		}
		int k = 0;
		for (int l = ao[0]; l != 0; l = next[l])
			k++;
		int[][] tr = new int[k][];
		k = 0;
		int dmax = 0;
		for (int l = ao[0]; l != 0; l = next[l]) {
			int j = jj[l];
			d_ = 0;
			dfs(0, j, k, 1);
			dmax = Math.max(dmax, d_);
			tr[k] = new int[d_ + 1];
			k++;
		}
		int[] tr_ = new int[dmax + 1];
		while (q-- > 0) {
			int t = sc.nextInt();
			int i = sc.nextInt() - 1;
			int h = hh[i];
			if (t == 0) {
				int x = sc.nextInt();
				int d = sc.nextInt();
				if (i == 0) {
					update(tr_, 0, x);
					update(tr_, d + 1, -x);
				} else if (d < dd[i]) {
					update(tr[h], dd[i] - d, x);
					update(tr[h], dd[i] + d + 1, -x);
				} else {
					update(tr_, 0, x);
					update(tr_, d - dd[i] + 1, -x);
					update(tr[h], d - dd[i] + 1, x);
					update(tr[h], dd[i] + d + 1, -x);
				}
			} else {
				int ans;
				if (i == 0)
					ans = query(tr_, 0);
				else
					ans = query(tr[h], dd[i]) + query(tr_, dd[i]);
				println(ans);
			}
		}
	}
}
