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;
vector<int> g[200009];
vector<vector<int> > u;
vector<vector<int> > T;
int L[200009], STree[4 * 200009], b[200009];
int n, m, a, mal, ch, v, x, d, t, maxlvl;
void dfs(int z, int par, int nah) {
  b[z] = nah;
  u[nah].push_back(z);
  L[z] = L[par] + 1;
  if (L[z] > maxlvl) maxlvl = L[z];
  for (int i = 0; i < g[z].size(); i++)
    if (g[z][i] != par) dfs(g[z][i], z, nah);
}
void art(int a, int b, int l, int r, int node, int x) {
  if (l > b || r < a) {
    return;
  }
  if (a <= l && r <= b) {
    STree[node] += x;
    return;
  }
  art(a, b, l, (l + r) / 2, node * 2, x);
  art(a, b, (l + r) / 2 + 1, r, node * 2 + 1, x);
}
void inc(int y, int a, int b, int l, int r, int node, int x) {
  if (l > b || r < a) {
    return;
  }
  if (a <= l && r <= b) {
    T[y][node] += x;
    return;
  }
  inc(y, a, b, l, (l + r) / 2, node * 2, x);
  inc(y, a, b, (l + r) / 2 + 1, r, node * 2 + 1, x);
}
int tap(int y, int z, int l, int r, int node) {
  if (l == r) return T[y][node];
  if (z <= (l + r) / 2)
    return tap(y, z, l, (l + r) / 2, node * 2) + T[y][node];
  else
    return tap(y, z, (l + r) / 2 + 1, r, node * 2 + 1) + T[y][node];
}
int goz(int y, int l, int r, int node) {
  if (l == r) return STree[node];
  if (y <= (l + r) / 2)
    return goz(y, l, (l + r) / 2, node * 2) + STree[node];
  else
    return goz(y, (l + r) / 2 + 1, r, node * 2 + 1) + STree[node];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &mal);
    g[a].push_back(mal);
    g[mal].push_back(a);
    if (a == 1 || mal == 1) ch++;
  }
  L[1] = 1;
  T.resize(ch);
  u.resize(ch);
  for (int i = 0; i < g[1].size(); i++) {
    u[i].push_back(1);
    dfs(g[1][i], 1, i);
    T[i].resize(int(u[i].size()) * 4 + 3);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &t);
    if (t == 0) {
      scanf("%d%d%d", &v, &x, &d);
      if (L[v] - 1 <= d) {
        art(1, min(d - L[v] + 2, maxlvl), 1, maxlvl, 1, x);
        if (d - L[v] + 3 <= maxlvl)
          inc(b[v], d - L[v] + 3, min(d + L[v], maxlvl), 1, u[b[v]].size(), 1,
              x);
      } else
        inc(b[v], L[v] - d, min(L[v] + d, int(u[b[v]].size())), 1,
            u[b[v]].size(), 1, x);
    } else {
      scanf("%d", &v);
      printf("%d\n",
             tap(b[v], L[v], 1, u[b[v]].size(), 1) + goz(L[v], 1, maxlvl, 1));
    }
  }
  return 0;
}
