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;
struct node {
  int Tn, g;
} P[100004];
vector<int> T[100004], col[100004], t0;
int n, num, M[100004];
void dfs(int x, int m, int f, int t) {
  P[x].Tn = m, P[x].g = t;
  col[num].push_back(0);
  for (int i = 0; i < T[x].size(); i++)
    if (T[x][i] != f) dfs(T[x][i], m, x, t + 1);
  if (t > M[0]) M[0] = t;
  return;
}
void PushDown(int M, int len, int now) {
  if (!col[M][now]) return;
  col[M][now << 1] += col[M][now];
  col[M][(now << 1) | 1] += col[M][now];
  col[M][now] = 0;
  return;
}
void update(int M, int L, int R, int c, int l, int r, int now) {
  if (L <= l && R >= r) {
    col[M][now] += c;
    return;
  }
  PushDown(M, r - l + 1, now);
  int mid = (l + r) >> 1;
  if (L <= mid) update(M, L, R, c, l, mid, now << 1);
  if (R > mid) update(M, L, R, c, mid + 1, r, (now << 1) | 1);
  return;
}
int query(int M, int p, int l, int r, int now) {
  if (l == r) return col[M][now];
  PushDown(M, r - l + 1, now);
  int mid = (l + r) >> 1;
  if (p <= mid) return query(M, p, l, mid, now << 1);
  if (p > mid) return query(M, p, mid + 1, r, (now << 1) | 1);
}
int main() {
  int i, q;
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; i++) T[i].clear();
  for (i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    T[x].push_back(y), T[y].push_back(x);
  }
  num = 0, P[1].g = 0, M[0] = 0;
  for (i = 0; i < T[1].size(); i++) {
    col[++num].clear(), col[num].push_back(0);
    dfs(T[1][i], num, 1, 1);
  }
  col[0].clear();
  i = M[0] << 2;
  while (i--) col[0].push_back(0);
  for (i = 1; i <= num; i++) {
    int x = col[i].size();
    M[i] = x - 1, x <<= 2;
    col[i].clear();
    while (x--) col[i].push_back(0);
  }
  while (q--) {
    int tp, v, x, d;
    scanf("%d", &tp);
    if (!tp) {
      int l, r, p, t;
      scanf("%d%d%d", &v, &x, &d);
      if (P[v].g - d <= 0) {
        l = d - P[v].g + 1, update(0, 0, min(M[0], l - 1), x, 0, M[0], 1);
      } else
        l = P[v].g - d;
      if (v == 1) continue;
      r = min(M[P[v].Tn], P[v].g + d);
      if (r < l) continue;
      update(P[v].Tn, l, r, x, 1, M[P[v].Tn], 1);
    } else {
      int ans = 0;
      scanf("%d", &v);
      ans = query(0, P[v].g, 0, M[0], 1);
      if (v != 1) ans += query(P[v].Tn, P[v].g, 1, M[P[v].Tn], 1);
      printf("%d\n", ans);
    }
  }
  return 0;
}
