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 = 111111;
struct bit {
  int s[MAXN];
  inline int getsum(int x) {
    int ret = 0;
    for (; x; x -= x & -x) ret += s[x];
    return ret;
  }
  inline void update(int x, int v) {
    for (; x < MAXN; x += x & -x) s[x] += v;
  }
} a, b;
int edgecnt, h[MAXN], p[MAXN << 1], nxt[MAXN << 1], c[MAXN], belong[MAXN],
    no[MAXN], size[MAXN];
inline void addedge(int u, int v) {
  p[++edgecnt] = v;
  nxt[edgecnt] = h[u];
  h[u] = edgecnt;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  c[1] = MAXN;
  int nodecnt = 0, listcnt = 0;
  for (int k = h[1]; k; k = nxt[k])
    if (!c[p[k]]) {
      ++listcnt;
      for (int t = k; t;)
        if (!c[p[t]]) {
          c[p[t]] = ++nodecnt;
          belong[p[t]] = listcnt;
          no[p[t]] = ++size[listcnt];
          t = h[p[t]];
        } else
          t = nxt[t];
    }
  int root = 0;
  while (q--) {
    int op, u, x, d;
    scanf("%d%d", &op, &u);
    if (op)
      printf("%d\n", u == 1 ? root : a.getsum(c[u]) + b.getsum(no[u]));
    else {
      scanf("%d%d", &x, &d);
      if (u == 1) {
        root += x;
        b.update(1, x);
        b.update(d + 1, -x);
      } else {
        if (d > no[u] - 1) {
          root += x;
          b.update(1, x);
          b.update(d - no[u] + 1, -x);
          a.update(c[u] - no[u] + 1, -x);
          a.update(min(c[u] + size[belong[u]] - no[u] + 1,
                       c[u] - no[u] + 1 + d - no[u]),
                   x);
        }
        a.update(c[u] - min(no[u] - 1, d), x);
        a.update(c[u] + min(d, size[belong[u]] - no[u]) + 1, -x);
      }
    }
  }
  return 0;
}
