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> edge[113111];
int n, q, st[113111][2], sc[113111];
long long a1[113111], a2[113111];
void dfs(int fr, int to, int th, int level) {
  st[to][0] = th;
  st[to][1] = level;
  for (int j = 0; j < edge[to].size(); j++) {
    if (edge[to][j] == fr) continue;
    dfs(to, edge[to][j], th, level + 1);
  }
}
void update(long long x, int wh, long long *a, int n) {
  if (wh <= 0) return;
  for (; wh <= n; wh += (wh) & (-wh)) a[wh] += x;
}
long long getsum(int wh, long long *a) {
  long long ret = 0;
  for (; wh > 0; wh -= (wh) & (-wh)) ret += a[wh];
  return ret;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  int son = edge[1].size();
  for (int i = 0; i < son; i++) dfs(1, edge[1][i], i + 1, 1);
  for (int i = 2; i <= n; i++) sc[st[i][0]]++;
  for (int i = 2; i <= son; i++) sc[i] += sc[i - 1];
  long long ans1 = 0;
  for (int i = 0; i < q; i++) {
    int qk;
    cin >> qk;
    if (qk == 0) {
      int v, x, d;
      cin >> v >> x >> d;
      if (v == 1) {
        update(x, min(d, n), a1, n);
        ans1 += x;
      } else if (d < st[v][1] - 1) {
        update(-x, st[v][1] - d - 1, a2 + sc[st[v][0] - 1],
               sc[st[v][0]] - sc[st[v][0] - 1]);
        update(x, min(st[v][1] + d, sc[st[v][0]] - sc[st[v][0] - 1]),
               a2 + sc[st[v][0] - 1], sc[st[v][0]] - sc[st[v][0] - 1]);
      } else {
        if (st[v][1] - d <= 0) ans1 += x;
        update(x, d - st[v][1], a1, n);
        update(-x, min(d - st[v][1], sc[st[v][0]] - sc[st[v][0] - 1]),
               a2 + sc[st[v][0] - 1], sc[st[v][0]] - sc[st[v][0] - 1]);
        update(x, min(st[v][1] + d, sc[st[v][0]] - sc[st[v][0] - 1]),
               a2 + sc[st[v][0] - 1], sc[st[v][0]] - sc[st[v][0] - 1]);
      }
    } else {
      int v;
      cin >> v;
      if (v == 1)
        cout << ans1 << endl;
      else {
        long long ans = 0;
        ans = getsum(n, a1) - getsum(st[v][1] - 1, a1);
        ans += getsum(sc[st[v][0]] - sc[st[v][0] - 1], a2 + sc[st[v][0] - 1]) -
               getsum(st[v][1] - 1, a2 + sc[st[v][0] - 1]);
        cout << ans << endl;
      }
    }
  }
  return 0;
}
