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 N = 100005;
int basearray[N], pz = -1;
int chainno = -1, chainhead[N], pos[N], chainnum[N], chainend[N];
int depth[N], n;
int st1[4 * N], st2[4 * N];
vector<int> adj[N];
void update1(int ss, int se, int ind, int val, int i) {
  if (ss > se || ind < ss || ind > se) return;
  if (ss == se) {
    st1[i] += val;
    return;
  }
  int mid = (ss + se) / 2;
  update1(ss, mid, ind, val, 2 * i + 1);
  update1(mid + 1, se, ind, val, 2 * i + 2);
  st1[i] = st1[2 * i + 1] + st1[2 * i + 2];
}
long long query1(int ss, int se, int l, int r, int i) {
  if (ss > se || ss > r || se < l) return 0;
  if (l <= ss && se <= r) {
    return st1[i];
  }
  int mid = (ss + se) / 2;
  return query1(ss, mid, l, r, 2 * i + 1) +
         query1(mid + 1, se, l, r, 2 * i + 2);
}
void update2(int ss, int se, int ind, int val, int i) {
  if (ss > se || ind < ss || ind > se) return;
  if (ss == se) {
    st2[i] += val;
    return;
  }
  int mid = (ss + se) / 2;
  update2(ss, mid, ind, val, 2 * i + 1);
  update2(mid + 1, se, ind, val, 2 * i + 2);
  st2[i] = st2[2 * i + 1] + st2[2 * i + 2];
}
long long query2(int ss, int se, int l, int r, int i) {
  if (ss > se || ss > r || se < l) return 0;
  if (l <= ss && se <= r) {
    return st2[i];
  }
  int mid = (ss + se) / 2;
  return query2(ss, mid, l, r, 2 * i + 1) +
         query2(mid + 1, se, l, r, 2 * i + 2);
}
void dfs(int s, int p) {
  depth[s] = depth[p] + 1;
  pos[s] = ++pz;
  chainnum[pz] = chainno;
  if (chainhead[chainno] == -1) chainhead[chainno] = pz;
  chainend[chainno] = pz;
  for (auto it : adj[s]) {
    if (it != p) dfs(it, s);
  }
}
void add(int v, int d, int x) {
  if (v == 0) {
    update2(0, n - 1, min(d, n - 1), x, 0);
    return;
  }
  if (pos[v] + d + 1 < n - 1 && chainnum[pos[v]] == chainnum[pos[v] + d + 1]) {
    update1(0, n - 1, pos[v] + d + 1, -x, 0);
  }
  int ch = chainhead[chainnum[pos[v]]];
  if (pos[v] - ch >= d) {
    update1(0, n - 1, pos[v] - d, x, 0);
  } else {
    int con = pos[v] - ch + 1;
    if (d - con + ch <= chainend[chainnum[pos[v]]]) {
      update1(0, n - 1, d - con + ch, x, 0);
    }
    int in = min(d - con, n - 1);
    update2(0, n - 1, in, x, 0);
  }
}
long long query(int v) {
  long long ans = 0;
  if (v != 0) {
    int ch = chainhead[chainnum[pos[v]]];
    ans = query1(0, n - 1, ch, pos[v], 0);
  }
  ans += query2(0, n - 1, depth[v], n - 1, 0);
  return ans;
}
int main() {
  int q, i;
  scanf("%d%d", &n, &q);
  for (i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    adj[x - 1].push_back(y - 1);
    adj[y - 1].push_back(x - 1);
  }
  memset(chainhead, -1, sizeof(chainhead));
  for (auto it : adj[0]) {
    chainno++;
    dfs(it, 0);
  }
  while (q--) {
    int type, v, d, x;
    scanf("%d", &type);
    if (type == 0) {
      scanf("%d%d%d", &v, &x, &d);
      add(v - 1, d, x);
    } else {
      scanf("%d", &v);
      printf("%I64d\n", query(v - 1));
    }
  }
}
