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 long long M = 1e6;
vector<int> G[100005];
vector<int> vc[100005];
int n, q, L[100005], maxn, ans0;
long long match[100005];
bool vis[100005];
int lowbit(int x) { return x & (-x); }
void update(int l, int pos, int x) {
  for (int i = l; i <= L[pos]; i += lowbit(i)) vc[pos][i] += x;
}
int query(int l, int pos) {
  int ans = 0;
  for (int i = l; i > 0; i -= lowbit(i)) ans += vc[pos][i];
  return ans;
}
void dfs(int x, int d, int l) {
  vc[d].push_back(0);
  vis[x] = 1;
  match[x] = 1ll * d * M + 1ll * l;
  for (int i = 0; i < G[x].size(); i++) {
    int tmp = G[x][i];
    if (!vis[tmp]) {
      dfs(tmp, d, l + 1);
    }
  }
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  int siz = G[1].size();
  vis[1] = 1;
  for (int i = 1; i <= siz; i++) {
    vc[i].push_back(0);
    dfs(G[1][i - 1], i, 1);
  }
  for (int i = 1; i <= siz; i++) {
    L[i] = vc[i].size() - 1;
    maxn = max(maxn, L[i]);
  }
  L[0] = maxn;
  for (int i = 0; i <= L[0]; i++) vc[0].push_back(0);
  while (q--) {
    int flag;
    scanf("%d", &flag);
    if (flag) {
      int u;
      scanf("%d", &u);
      int pos = match[u] / M;
      int l = match[u] % M;
      if (!l)
        printf("%d\n", ans0);
      else
        printf("%d\n", query(l, pos) + query(l, 0));
    } else {
      int v, x, d;
      scanf("%d %d %d", &v, &x, &d);
      if (v == 1) {
        ans0 += x;
        update(1, 0, x);
        update(min(maxn, d) + 1, 0, -x);
        continue;
      }
      int pos = match[v] / M;
      int l = match[v] % M;
      int r1 = min(l + d, L[pos]);
      int l1 = max(l - d, 1);
      update(l1, pos, x);
      update(r1 + 1, pos, -x);
      if (l - d <= 0) ans0 += x;
      if (l - d < 0) {
        update(1, 0, x);
        update(d - l + 1, 0, -x);
        update(1, pos, -x);
        update(d - l + 1, pos, x);
      }
    }
  }
  return 0;
}
